Equivalent Algebraic Expressions Alıştırma Soruları — SAT — Sayfa 3

Soru 41Soru →

For all $x > 0$ , the expression $\left( \frac{x^{4/3} + 4x^{2/3} + 16}{x^{2/3} + 2x^{1/3} + 4} + 2x^{1/3} \right)^3 - x^2 - 12x^{4/3} - 48x^{2/3}$ is equivalent to a constant $C$ . What is the value of $C$ ?

Cevabı ve açıklamayı göster

Cevap: 64

Cevap

The constant value is 64.

The expression inside the parentheses simplifies to

x^{2/3} + 4

after factoring the numerator as

(x^{2/3} + 2x^{1/3} + 4)(x^{2/3} - 2x^{1/3} + 4)

and canceling the common factor in the denominator. Cubing

x^{2/3} + 4

yields

x^2 + 12x^{4/3} + 48x^{2/3} + 64

. Subtracting the remaining terms

x^2 + 12x^{4/3} + 48x^{2/3}

from this expansion results in the constant value 64.

Adım Adım Çözüm

1

Substitute

u = x^{1/3}

into the expression to simplify the fractional exponents.

The terms become

x^{1/3} = u

,

x^{2/3} = u^2

,

x^{4/3} = u^4

, and

x^2 = u^6

. The expression inside the parentheses is rewritten as

\frac{u^4 + 4u^2 + 16}{u^2 + 2u + 4} + 2u

.

Using a temporary variable

u

simplifies the algebraic factoring and manipulation of terms with fractional exponents.

2

Factor the numerator

u^4 + 4u^2 + 16

by completing the square.

u^4 + 4u^2 + 16 = (u^2 + 4)^2 - 4u^2 = (u^2 + 2u + 4)(u^2 - 2u + 4)

.

Expressing the quartic polynomial as a difference of squares allows it to be factored into two quadratic polynomials.

3

Simplify the rational expression and add

2u

.

\frac{(u^2 + 2u + 4)(u^2 - 2u + 4)}{u^2 + 2u + 4} + 2u = (u^2 - 2u + 4) + 2u = u^2 + 4

.

Canceling the common factor

u^2 + 2u + 4

in the numerator and denominator simplifies the expression inside the parentheses to

u^2 + 4

.

4

Substitute

u = x^{1/3}

back into

u^2 + 4

and cube the expression.

(x^{2/3} + 4)^3 = (x^{2/3})^3 + 3(x^{2/3})^2(4) + 3(x^{2/3})(16) + 64 = x^2 + 12x^{4/3} + 48x^{2/3} + 64

.

Applying the binomial expansion formula

(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3

expands the cubed expression.

5

Subtract the remaining terms from the expanded expression.

(x^2 + 12x^{4/3} + 48x^{2/3} + 64) - x^2 - 12x^{4/3} - 48x^{2/3} = 64

.

Subtracting the variable terms cancels them out entirely, leaving the constant value 64.

Anahtar Kavram

Factoring quartic polynomials using the difference of squares and simplifying rational expressions with fractional exponents.

Soru 42Soru →

For all $x > 2$ , which of the following is equivalent to the expression $\frac{3x^2 - 5x - 2}{x^2 - 4} - \frac{x - 1}{x + 2}$ ?

\frac{2x}{x + 2}

\frac{2x + 2}{x + 2}

\frac{2x + 2}{x - 2}

\frac{x + 1}{x + 2}

Cevabı ve açıklamayı göster

Cevap: $\frac{2x + 2}{x + 2}$

Cevap

The expression is equivalent to

\frac{2x + 2}{x + 2}

.

The correct answer is obtained by first factoring the first term:

\frac{3x^2 - 5x - 2}{x^2 - 4} = \frac{(3x + 1)(x - 2)}{(x - 2)(x + 2)}

. Canceling the common factor of

x - 2

yields

\frac{3x + 1}{x + 2}

. Subtracting the second term gives

\frac{3x + 1 - (x - 1)}{x + 2} = \frac{3x + 1 - x + 1}{x + 2} = \frac{2x + 2}{x + 2}

.

Adım Adım Çözüm

1

Factor the numerator and the denominator of the first term of the expression.

The numerator factors as

3x^2 - 5x - 2 = (3x + 1)(x - 2)

, and the denominator factors as

x^2 - 4 = (x - 2)(x + 2)

.

This allows common factors in the numerator and denominator to be identified and canceled.

2

Simplify the first term by canceling the common factor

(x - 2)

for

x > 2

.

The first term simplifies to

\frac{3x + 1}{x + 2}

.

For

x > 2

,

x - 2 \neq 0

, so we can divide both numerator and denominator by

x - 2

to simplify the fraction.

3

Subtract the second term from the simplified first term.

The expression becomes

\frac{3x + 1}{x + 2} - \frac{x - 1}{x + 2} = \frac{(3x + 1) - (x - 1)}{x + 2}

.

Since both fractions have the same denominator,

x + 2

, their numerators can be subtracted directly.

4

Distribute the negative sign in the numerator and combine like terms.

\frac{3x + 1 - x + 1}{x + 2} = \frac{2x + 2}{x + 2}

.

Distributing the subtraction to both terms in the parenthesis

(x - 1)

gives

-x + 1

. Combining

3x - x

yields

2x

, and

1 + 1

yields

2

.

Anahtar Kavram

Simplifying rational expressions by factoring and performing algebraic operations with common denominators.

Tahmini Süre:1m 30s

Soru 43Soru →

For all $x > 0$ and $y > 0$ , the expression $\frac{(x^2y^3)^a}{(xy^2)^2}$ is equivalent to $x^6y^8$ , where $a$ is a constant. What is the value of $a$ ?

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Cevap: 4

Cevap

4

Applying the rules of exponents, the expression

\frac{(x^2y^3)^a}{(xy^2)^2}

simplifies to

\frac{x^{2a}y^{3a}}{x^2y^4} = x^{2a-2}y^{3a-4}

. Setting this equal to the equivalent expression

x^6y^8

gives the system of equations

2a - 2 = 6

and

3a - 4 = 8

. Solving either equation yields

a = 4

.

Adım Adım Çözüm

1

Apply the power rule of exponents to the numerator and denominator.

Numerator:

x^{2a}y^{3a}

, Denominator:

x^2y^4

To expand the parentheses by multiplying the outer exponent with the inner exponents.

2

Apply the quotient rule of exponents to divide the numerator by the denominator.

x^{2a-2}y^{3a-4}

To simplify the rational expression by subtracting the exponents in the denominator from the exponents in the numerator.

3

Equate the simplified exponent of

x

to the exponent of

x

in the given equivalent expression.

2a - 2 = 6

Equivalent expressions must have identical exponents for corresponding variable bases.

4

Solve the linear equation for

a

.

a = 4

To find the constant value that satisfies the equation.

5

Verify the value of

a

using the exponents of

y

.

3(4) - 4 = 8

, which is true.

To ensure consistency across both variable exponents.

Anahtar Kavram

Equivalent algebraic expressions involving exponent rules

Soru 44Soru →

For all $x > 0$ and $x \neq 1$ , the expression $\frac{x^{5/2} - x^{3/2} - x + 1}{x^{3/2} - x - x^{1/2} + 1}$ is equivalent to $x + x^a + b$ , where $a$ and $b$ are constants and $a < 1$ . What is the value of $2a + b$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The correct answer is 2.

Substituting

u = x^{1/2}

transforms the expression into

\frac{u^5 - u^3 - u^2 + 1}{u^3 - u^2 - u + 1}

. Factoring the numerator by grouping yields

(u^3 - 1)(u^2 - 1)

, and factoring the denominator yields

(u^2 - 1)(u - 1)

. Canceling the common factor

(u^2 - 1)

leaves

\frac{u^3 - 1}{u - 1}

. Applying the difference of cubes formula to factor

u^3 - 1

as

(u - 1)(u^2 + u + 1)

allows us to cancel the

(u - 1)

term, leaving

u^2 + u + 1

. Re-substituting

u = x^{1/2}

results in the equivalent expression

x + x^{1/2} + 1

. Comparing this to

x + x^a + b

with

a < 1

gives

a = 0.5

and

b = 1

. Evaluating

2a + b

yields

2(0.5) + 1 = 2

.

Adım Adım Çözüm

1

Introduce a variable substitution to eliminate fractional exponents.

Let

u = x^{1/2}

, which implies

x = u^2

. The expression becomes

\frac{u^5 - u^3 - u^2 + 1}{u^3 - u^2 - u + 1}

.

Substitution simplifies working with fractional exponents and makes factoring patterns easier to identify.

2

Factor the numerator by grouping terms.

The numerator factors as

u^3(u^2 - 1) - (u^2 - 1) = (u^3 - 1)(u^2 - 1)

.

Grouping common terms helps simplify polynomials into products of lower-degree factors.

3

Factor the denominator by grouping terms.

The denominator factors as

u^2(u - 1) - (u - 1) = (u^2 - 1)(u - 1)

.

Factoring the denominator reveals common terms that can be canceled with the numerator.

4

Combine the factored forms and cancel the common factor.

The expression becomes

\frac{(u^3 - 1)(u^2 - 1)}{(u^2 - 1)(u - 1)} = \frac{u^3 - 1}{u - 1}

for

u \neq 1

.

Canceling

(u^2 - 1)

simplifies the rational expression since

x \neq 1

ensures

u^2 - 1 \neq 0

.

5

Simplify the remaining rational expression using the difference of cubes formula.

Factoring

u^3 - 1

as

(u - 1)(u^2 + u + 1)

and canceling

(u - 1)

yields

u^2 + u + 1

.

The difference of cubes formula allows cancellation of the linear term in the denominator.

6

Substitute back the original variable and determine the constants.

Substituting

u = x^{1/2}

back gives

x + x^{1/2} + 1

, meaning

a = 0.5

and

b = 1

.

This puts the expression in the final requested form

x + x^a + b

to find

a

and

b

.

7

Calculate the requested value

2a + b

.

2(0.5) + 1 = 2

.

This computes the final numeric answer requested by the question.

Anahtar Kavram

Simplifying rational expressions with fractional exponents by substitution and grouping

Soru 45Soru →

For all $x > 5$ , which of the following is equivalent to the expression $\frac{3x^2 - 14x - 5}{3x + 1} - \frac{x^2 - 25}{x - 5}$ ?

-10

0

2x

2x - 10

Cevabı ve açıklamayı göster

Cevap: $-10$

Cevap

-10

Factoring the numerators allows both rational expressions to be simplified. The first expression,

\frac{3x^2 - 14x - 5}{3x + 1}

, factors as

\frac{(3x + 1)(x - 5)}{3x + 1}

, which simplifies to

x - 5

since

x > 5

. The second expression,

\frac{x^2 - 25}{x - 5}

, factors as

\frac{(x - 5)(x + 5)}{x - 5}

, which simplifies to

x + 5

. Subtracting the second simplified expression from the first gives

(x - 5) - (x + 5) = x - 5 - x - 5 = -10

. Therefore, the equivalent expression is

-10

.

Adım Adım Çözüm

1

Factor the numerator of the first rational expression,

3x^2 - 14x - 5

.

(3x + 1)(x - 5)

To find common factors that can be simplified with the denominator

3x + 1

.

2

Simplify the first expression,

\frac{3x^2 - 14x - 5}{3x + 1}

, for

x > 5

.

x - 5

Since

x > 5

,

3x + 1 \neq 0

, allowing us to divide out the common factor

(3x + 1)

.

3

Factor the numerator of the second rational expression,

x^2 - 25

.

(x - 5)(x + 5)

Using the difference of squares identity,

a^2 - b^2 = (a - b)(a + b)

.

4

Simplify the second expression,

\frac{x^2 - 25}{x - 5}

, for

x > 5

.

x + 5

Since

x > 5

,

x - 5 \neq 0

, allowing us to divide out the common factor

(x - 5)

.

5

Subtract the second simplified expression from the first.

(x - 5) - (x + 5) = x - 5 - x - 5 = -10

Distributing the negative sign through the parentheses and combining like terms yields the final simplified equivalent value.

Anahtar Kavram

Simplifying rational expressions by factoring and performing polynomial operations.

Soru 46Soru →

For all $x > 0$ , which of the following is equivalent to the expression $\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 \left(x + 1 + \frac{1}{x}\right)$ ?

\frac{(x+1)(x^3-1)}{x^2}

\frac{(x^2+1)(x^2+x+1)}{x^2}

\frac{(x-1)(x^3-1)}{x^2}

\frac{(x-1)(x^3+1)}{x^2}

Cevabı ve açıklamayı göster

Cevap: $\frac{(x-1)(x^3-1)}{x^2}$

Cevap

\frac{(x-1)(x^3-1)}{x^2}

The correct expression is obtained by first converting

\sqrt{x} - \frac{1}{\sqrt{x}}

to

\frac{x-1}{\sqrt{x}}

, and then squaring it to get

\frac{(x-1)^2}{x}

. Next, the term

x + 1 + \frac{1}{x}

is converted to

\frac{x^2+x+1}{x}

. Multiplying these two results yields

\frac{(x-1)^2(x^2+x+1)}{x^2}

. Grouping

(x-1)(x^2+x+1)

and substituting it with the difference of cubes identity,

x^3-1

, gives the final equivalent expression

\frac{(x-1)(x^3-1)}{x^2}

.

Adım Adım Çözüm

1

Simplify the first factor by writing

\sqrt{x} - \frac{1}{\sqrt{x}}

under a common denominator and squaring it.

\left(\frac{x-1}{\sqrt{x}}\right)^2 = \frac{(x-1)^2}{x}

This allows us to combine the terms in the first factor before multiplication.

2

Write the second factor

x + 1 + \frac{1}{x}

under a common denominator.

\frac{x^2+x+1}{x}

This allows us to multiply the two fractional factors easily.

3

Multiply the two simplified factors together.

\frac{(x-1)^2(x^2+x+1)}{x^2} = \frac{(x-1)(x-1)(x^2+x+1)}{x^2}

Multiplying the numerators and denominators of the fractions gives the product of the terms.

4

Apply the difference of cubes factorization formula:

(x-1)(x^2+x+1) = x^3-1

.

\frac{(x-1)(x^3-1)}{x^2}

This simplifies the product to its final equivalent form.

Anahtar Kavram

Simplifying rational expressions and applying algebraic identities (specifically the binomial square and difference of cubes).

Alternatif Yöntem

Alternatively, you can substitute a convenient value for

x

, such as

x = 4

, into the original expression and each of the options, then compare the results.

Tahmini Süre:2m 0s

Soru 47Soru →

For all $x > 3$ , which of the following expressions is equivalent to $\frac{x^2 - 9}{x^2 - 2x - 3} \cdot \frac{x + 1}{x}$ ?

\frac{x+3}{x+1}

\frac{x+3}{x}

\frac{x-3}{x}

\frac{x^2+4x+3}{x^2-x}

Cevabı ve açıklamayı göster

Cevap: $\frac{x+3}{x}$

Cevap

\frac{x+3}{x}

Factoring the numerator of the first fraction as

(x-3)(x+3)

and its denominator as

(x-3)(x+1)

allows the common factor

(x-3)

to cancel out, leaving

\frac{x+3}{x+1}

. Multiplying this result by the second fraction,

\frac{x+1}{x}

, permits the cancellation of the common factor

(x+1)

, which simplifies the entire expression to the equivalent form

\frac{x+3}{x}

.

Adım Adım Çözüm

1

Factor the numerator and the denominator of the first rational expression:

\frac{x^2 - 9}{x^2 - 2x - 3}

.

The numerator factors as

(x - 3)(x + 3)

using the difference of squares identity. The denominator factors as

(x - 3)(x + 1)

by finding two numbers that multiply to

-3

and add to

-2

.

Factoring polynomials is necessary to identify and cancel common factors.

2

Simplify the first fraction by canceling the common factor

(x-3)

from both the numerator and denominator.

The first fraction simplifies to

\frac{x+3}{x+1}

.

Since

x > 3

, the term

x-3

is non-zero, making division by it valid.

3

Multiply the simplified first fraction by the second fraction:

\frac{x+3}{x+1} \cdot \frac{x+1}{x}

.

The common factor

(x+1)

in the numerator and denominator cancels out, resulting in

\frac{x+3}{x}

.

Canceling the common factor

(x+1)

yields the simplest equivalent expression.

Anahtar Kavram

Simplifying rational expressions by factoring polynomials and canceling common factors.

Soru 48Soru →

For all $x > 3$ , the expression $\frac{x^3 - 2x^2 - 9x + 18}{x^2 - 5x + 6} - \frac{x^3 + 3x^2 - 4x - 12}{x^2 + 5x + 6}$ is equivalent to the constant $k$ . What is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of the constant

k

is 5.

Factoring the numerators by grouping yields

x^3 - 2x^2 - 9x + 18 = (x-3)(x+3)(x-2)

and

x^3 + 3x^2 - 4x - 12 = (x-2)(x+2)(x+3)

. Dividing out their respective denominators

(x-2)(x-3)

and

(x+2)(x+3)

leaves the simplified linear expressions

x+3

and

x-2

. Subtracting these yields

(x+3) - (x-2) = 5

.

Adım Adım Çözüm

1

Factor the numerator and denominator of the first rational expression and cancel common factors.

x + 3

To simplify the first fraction by reducing it to its lowest terms.

2

Factor the numerator and denominator of the second rational expression and cancel common factors.

x - 2

To simplify the second fraction by reducing it to its lowest terms.

3

Subtract the second simplified expression from the first simplified expression.

5

To find the constant value equivalent to the given difference of rational expressions, ensuring to distribute the negative sign to all parts of the subtracted binomial.

Anahtar Kavram

Simplifying rational expressions by factoring cubic polynomials by grouping and quadratic trinomials

Soru 49Soru →

If the expression $\frac{6x^2 + 7x - 20}{2x - 3}$ is equivalent to $ax + b + \frac{c}{2x-3}$ for all $x \neq 1.5$ , where $a$ , $b$ , and $c$ are constants, what is the value of $a + b + c$ ?

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

The value of

a + b + c

is 15.

By dividing the numerator

6x^2 + 7x - 20

by the denominator

2x - 3

, we find that the quotient is

3x + 8

and the remainder is

4

. Thus, the expression can be rewritten as

3x + 8 + \frac{4}{2x-3}

. Comparing this to the given expression

ax + b + \frac{c}{2x-3}

, we obtain

a = 3

,

b = 8

, and

c = 4

. Their sum is

3 + 8 + 4 = 15

.

Adım Adım Çözüm

1

Set up the polynomial division of the numerator

6x^2 + 7x - 20

by the denominator

2x - 3

.

Dividing

6x^2 + 7x - 20

by

2x - 3

.

To express the rational expression in terms of a polynomial quotient and a remainder.

2

Divide the first term of the numerator by the first term of the denominator to determine the first quotient term.

The first term is

3x

. Subtracting

3x(2x - 3)

from the numerator leaves

16x - 20

.

\frac{6x^2}{2x} = 3x

, and subtracting

6x^2 - 9x

from the polynomial leaves the next term to be divided.

3

Divide the leading term of the remaining expression by the leading term of the denominator to determine the constant term of the quotient.

The constant term is

8

. Subtracting

8(2x - 3)

from

16x - 20

leaves a remainder of

4

.

\frac{16x}{2x} = 8

, and subtracting

16x - 24

from

16x - 20

gives the final constant remainder.

4

Compare the quotient and remainder to the given form to identify

a

,

b

, and

c

.

a = 3

,

b = 8

, and

c = 4

.

The quotient is

3x + 8

and the remainder is

4

, which matches the form

ax + b + \frac{c}{2x-3}

.

5

Calculate the sum of

a

,

b

, and

c

.

3 + 8 + 4 = 15

.

The question asks for the value of

a + b + c

.

Anahtar Kavram

Polynomial division and rewrite of rational expressions

Soru 50Soru →

For all $x > 2$ , which of the following is equivalent to the expression $\frac{3x^2 - 12}{x^2 - 2x} - \frac{2x^2 + 2x - 12}{x^2 + 3x}$ ?

\frac{x+2}{x}

\frac{x+10}{x}

\frac{x-12}{x}

\frac{x+14}{x}

Cevabı ve açıklamayı göster

Cevap: $\frac{x+10}{x}$

Cevap

\frac{x+10}{x}

The correct answer is obtained by factoring both rational expressions and simplifying them before subtracting. The first term factors into

\frac{3(x-2)(x+2)}{x(x-2)}

, which simplifies to

\frac{3(x+2)}{x}

. The second term factors into

\frac{2(x+3)(x-2)}{x(x+3)}

, which simplifies to

\frac{2(x-2)}{x}

. Subtracting these two expressions yields

\frac{3(x+2) - 2(x-2)}{x} = \frac{3x+6-2x+4}{x} = \frac{x+10}{x}

.

Adım Adım Çözüm

1

Factor the numerators and denominators of both rational expressions.

The expression becomes

\frac{3(x-2)(x+2)}{x(x-2)} - \frac{2(x+3)(x-2)}{x(x+3)}

.

Factoring helps identify common binomial factors in the numerator and denominator.

2

Simplify both terms by canceling common factors.

For

x > 2

, the expression simplifies to

\frac{3(x+2)}{x} - \frac{2(x-2)}{x}

.

Since

x > 2

, the terms

x-2

and

x+3

are non-zero and can be canceled.

3

Combine the simplified terms over the common denominator

x

.

\frac{3(x+2) - 2(x-2)}{x} = \frac{3x + 6 - (2x - 4)}{x}

Both terms share the common denominator

x

.

4

Distribute the negative sign in the numerator and combine like terms.

\frac{3x + 6 - 2x + 4}{x} = \frac{x+10}{x}

Distributing subtraction to the terms inside

(2x-4)

yields

-2x + 4

.

Anahtar Kavram

Simplifying and subtracting rational expressions by factoring and finding a common denominator

Soru 51Soru →

Which of the following is equivalent to the expression $\frac{3x^2 - 5x - 4}{x - 2}$ for all $x \neq 2$ ?

3x - 1 - \frac{6}{x - 2}

3x + 1 - \frac{2}{x - 2}

3x + 1 + \frac{2}{x - 2}

3x - 5 - \frac{4}{x - 2}

Cevabı ve açıklamayı göster

Cevap: $3x + 1 - \frac{2}{x - 2}$

Cevap

3x + 1 - \frac{2}{x - 2}

The correct answer represents the equivalent expression obtained by performing polynomial division on the rational expression. Dividing the numerator

3x^2 - 5x - 4

by the denominator

x - 2

yields a quotient of

3x + 1

and a remainder of

-2

. This can be written in the form of the quotient plus the remainder over the divisor, resulting in

3x + 1 - \frac{2}{x - 2}

.

Adım Adım Çözüm

1

Divide the leading term of the numerator,

3x^2

, by the leading term of the denominator,

x

.

The first term of the quotient is

3x

.

This starts the polynomial long division process.

2

Multiply

3x

by the divisor

(x - 2)

and subtract the result from the numerator.

(3x^2 - 5x - 4) - (3x^2 - 6x) = x - 4

.

Subtracting the multiplied term helps find the remainder of the first division step.

3

Divide the leading term of the remaining expression,

x

, by the leading term of the divisor,

x

.

The second term of the quotient is

1

.

To continue the division process with the remaining terms.

4

Multiply

1

by the divisor

(x - 2)

and subtract the result from

x - 4

.

(x - 4) - (x - 2) = -2

.

This step determines the final remainder of

-2

because the degree of the remainder is now less than the degree of the divisor.

5

Express the final result as the sum of the quotient and the remainder divided by the divisor.

3x + 1 - \frac{2}{x - 2}

To construct the equivalent algebraic expression.

Anahtar Kavram

Equivalent Algebraic Expressions

Soru 52Soru →

For all $x > 1$ , the expression $\frac{2x^2 + 7x - 4}{x^2 - 1} \cdot \frac{x - 1}{2x - 1}$ is equivalent to $\frac{x+k}{x+1}$ , where $k$ is a constant. What is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of the constant

k

is 4.

Factoring the numerator

2x^2 + 7x - 4

yields

(2x - 1)(x + 4)

and factoring the denominator

x^2 - 1

yields

(x - 1)(x + 1)

. Substituting these factored forms into the given product gives

\frac{(2x - 1)(x + 4)}{(x - 1)(x + 1)} \cdot \frac{x - 1}{2x - 1}

. Canceling the common factors

(2x - 1)

and

(x - 1)

simplifies the expression to

\frac{x + 4}{x + 1}

. Comparing this to

\frac{x + k}{x + 1}

shows that

k = 4

.

Adım Adım Çözüm

1

Factor the quadratic expression in the numerator:

2x^2 + 7x - 4

.

(2x - 1)(x + 4)

Factoring the numerator helps identify common factors that can be simplified.

2

Factor the difference of squares in the denominator:

x^2 - 1

.

(x - 1)(x + 1)

Factoring the denominator helps identify common factors that can be simplified.

3

Multiply the rational expressions and cancel out the common factors.

\frac{x + 4}{x + 1}

Since

x > 1

, the terms

(2x - 1)

and

(x - 1)

are not equal to zero and can be canceled.

4

Compare the resulting expression with

\frac{x + k}{x + 1}

to find the value of

k

.

k = 4

By matching the numerators of the equivalent expressions,

x + 4 = x + k

, which gives

k = 4

.

Anahtar Kavram

Factoring and simplifying products of rational expressions

Soru 53Öğrencilerin %0'i bunu doğru yanıtladıSoru →

For all $x > 3$ , which of the following is equivalent to the expression $\frac{2x^2 - 5x - 3}{x^2 - 9} - \frac{x}{x+3}$ ?

\frac{x + 1}{x + 3}

\frac{x - 1}{x + 3}

\frac{3x + 1}{x + 3}

\frac{2x^2 - 6x - 3}{x^2 - 9}

Cevabı ve açıklamayı göster

Cevap: $\frac{x + 1}{x + 3}$

Cevap

\frac{x + 1}{x + 3}

To simplify the expression, we factor the first term:

\frac{2x^2 - 5x - 3}{x^2 - 9} = \frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)}

. Canceling the common factor of

x - 3

leaves

\frac{2x + 1}{x + 3}

. Subtracting the second term gives

\frac{2x + 1}{x + 3} - \frac{x}{x + 3} = \frac{2x + 1 - x}{x + 3} = \frac{x + 1}{x + 3}

. This matches the correct expression.

Adım Adım Çözüm

1

Factor the numerator and denominator of the first rational expression.

The numerator

2x^2 - 5x - 3

factors into

(2x + 1)(x - 3)

. The denominator

x^2 - 9

is a difference of squares and factors into

(x - 3)(x + 3)

.

Factoring allows for the simplification of the rational expression by identifying common factors in the numerator and denominator.

2

Simplify the first rational expression by canceling the common factor.

\frac{(2x + 1)(x - 3)}{(x - 3)(x + 3)} = \frac{2x + 1}{x + 3}

for all

x > 3

.

Since

x > 3

, the term

x - 3

is non-zero, allowing us to divide both the numerator and denominator by

x - 3

.

3

Subtract the second expression from the simplified first expression.

\frac{2x + 1}{x + 3} - \frac{x}{x + 3} = \frac{2x + 1 - x}{x + 3} = \frac{x + 1}{x + 3}

Since both fractions now have the same denominator,

x + 3

, we subtract their numerators.

Anahtar Kavram

Simplifying rational expressions by factoring polynomials and performing operations on fractions with common denominators.

Tahmini Süre:1m 30s

Soru 54Soru →

If $x$ and $y$ are positive numbers, which of the following is equivalent to the expression $\frac{\sqrt{36x^5 y^3}}{3x(xy^3)^{1/2}}$ ?

2x\sqrt{x}

2x

12x

2xy

Cevabı ve açıklamayı göster

Cevap: $2x$

Cevap

The expression is equivalent to

2x

.

To find the equivalent expression, simplify the numerator and the denominator separately. The numerator

\sqrt{36x^5 y^3}

simplifies to

6x^{5/2}y^{3/2}

. The denominator

3x(xy^3)^{1/2}

simplifies to

3x^{3/2}y^{3/2}

. Dividing the simplified numerator by the simplified denominator yields

\frac{6x^{5/2}y^{3/2}}{3x^{3/2}y^{3/2}} = 2x

.

Adım Adım Çözüm

1

Convert the radical expression in the numerator to an expression with fractional exponents.

\sqrt{36x^5 y^3} = (36x^5 y^3)^{1/2} = 6x^{5/2}y^{3/2}

The square root of a product is the product of the square roots of each factor, where

\sqrt{a} = a^{1/2}

.

2

Apply exponent rules to simplify the denominator.

3x(xy^3)^{1/2} = 3x^1 \cdot x^{1/2} y^{3/2} = 3x^{3/2}y^{3/2}

When multiplying terms with the same base, add their exponents (

x^a \cdot x^b = x^{a+b}

).

3

Divide the simplified numerator by the simplified denominator.

\frac{6x^{5/2}y^{3/2}}{3x^{3/2}y^{3/2}} = 2x^{5/2 - 3/2}y^{3/2 - 3/2} = 2x^1 y^0 = 2x

When dividing terms with the same base, subtract the exponent of the denominator from the exponent of the numerator (

\frac{x^a}{x^b} = x^{a-b}

).

Anahtar Kavram

Equivalent Algebraic Expressions

Tahmini Süre:1m 15s

Soru 55Soru →

Which of the following expressions is equivalent to $\frac{x^2 - 4x - 12}{2x - 12} - \frac{x}{4}$ for any real number $x > 6$ ?

\frac{x - 4}{4}

\frac{x + 2}{4}

\frac{x + 4}{4}

\frac{3x + 4}{4}

Cevabı ve açıklamayı göster

Cevap: $\frac{x + 4}{4}$

Cevap

The expression stating that the equivalent form is the fraction with

x+4

in the numerator and

4

in the denominator

The correct equivalent expression is found by first factoring the rational expression

\frac{x^2 - 4x - 12}{2x - 12}

into

\frac{(x - 6)(x + 2)}{2(x - 6)}

. Since

x > 6

, the factor

x - 6

is non-zero and can be canceled, leaving

\frac{x + 2}{2}

. Converting

\frac{x + 2}{2}

to have a common denominator of

4

yields

\frac{2x + 4}{4}

. Finally, subtracting

\frac{x}{4}

gives the simplified result of

\frac{x + 4}{4}

.

Adım Adım Çözüm

1

Factor the quadratic numerator and the linear denominator of the first rational term.

\frac{x^2 - 4x - 12}{2x - 12} = \frac{(x - 6)(x + 2)}{2(x - 6)}

To identify and cancel common algebraic factors in the fraction.

2

Cancel the common factor

x - 6

from the numerator and denominator.

\frac{x + 2}{2}

Since

x > 6

, the term

x - 6

is non-zero and can be canceled.

3

Find a common denominator of

4

to perform subtraction with the second term,

-\frac{x}{4}

.

\frac{2(x + 2)}{4} - \frac{x}{4} = \frac{2x + 4}{4} - \frac{x}{4}

To subtract fractions, their denominators must be identical.

4

Subtract the numerators and combine terms over the common denominator.

\frac{2x + 4 - x}{4} = \frac{x + 4}{4}

Subtracting like terms in the numerator yields the final simplified expression.

Anahtar Kavram

Equivalent Algebraic Expressions

Alternatif Yöntem

Instead of simplifying algebraically, you can substitute a value for

x

that is greater than

6

. For example, if

x = 8

, the original expression evaluates to

\frac{8^2 - 4(8) - 12}{2(8) - 12} - \frac{8}{4} = \frac{20}{4} - 2 = 3

. Evaluating the correct option with

x = 8

yields

\frac{8+4}{4} = 3

, confirming equivalence.

Tahmini Süre:1m 30s

Soru 56Soru →

For all $x > 3$ , the expression $\frac{2x^2 - 11x + 15}{x - 3} + \frac{3x^2 - 12}{x - 2}$ is equivalent to $ax + b$ , where $a$ and $b$ are constants. What is the value of $a + b$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The value of

a + b

is

6

.

The correct answer is

6

. By factoring the numerators of both rational expressions, we can simplify them. The first term is

\frac{(2x-5)(x-3)}{x-3} = 2x-5

. The second term is

\frac{3(x-2)(x+2)}{x-2} = 3(x+2) = 3x+6

. Adding these simplified terms gives

(2x-5) + (3x+6) = 5x+1

. Comparing this to

ax+b

yields

a=5

and

b=1

, so

a+b = 5+1=6

.

Adım Adım Çözüm

1

Factor the numerator of the first rational expression and simplify.

\frac{2x^2 - 11x + 15}{x - 3} = 2x - 5

Since

2x^2 - 11x + 15 = (2x - 5)(x - 3)

, the factor

(x - 3)

divides out for all

x > 3

.

2

Factor the numerator of the second rational expression and simplify.

\frac{3x^2 - 12}{x - 2} = 3x + 6

Since

3x^2 - 12 = 3(x - 2)(x + 2)

, the factor

(x - 2)

divides out for all

x > 3

.

3

Combine the simplified terms by adding them.

5x + 1

Adding

(2x - 5)

and

(3x + 6)

yields

(2x + 3x) + (-5 + 6) = 5x + 1

.

4

Identify the values of

a

and

b

and calculate their sum.

a = 5

,

b = 1

, and

a + b = 6

Comparing

5x + 1

with

ax + b

gives

a = 5

and

b = 1

, so

a + b = 5 + 1 = 6

.

Anahtar Kavram

Simplifying rational expressions by factoring the numerator and dividing out common factors.

Soru 57Soru →

For all positive real numbers $x$ and $y$ , the expression $\frac{(4x^3 y^2)^2}{\sqrt{16x^8 y^6}}$ can be written in the equivalent form $ax^b y^c$ , where $a$ , $b$ , and $c$ are positive constants. What is the value of $a + b + c$ ?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

7

The expression simplifies to

4x^2 y^1

by applying the power of a product rule to the numerator to get

16x^6 y^4

, and simplifying the radical in the denominator to get

4x^4 y^3

. Dividing the terms yields

4x^2 y^1

. Identifying the coefficients and exponents gives

a = 4

,

b = 2

, and

c = 1

. The sum of these values is

4 + 2 + 1 = 7

.

Adım Adım Çözüm

1

Simplify the numerator of the expression.

(4x^3 y^2)^2 = 16x^6 y^4

Apply the power of a product rule

(ab)^n = a^n b^n

and the power of a power rule

(a^m)^n = a^{mn}

.

2

Simplify the denominator of the expression.

\sqrt{16x^8 y^6} = 4x^4 y^3

Take the square root of the coefficient and apply fractional exponent rules for positive variables:

\sqrt{16} = 4

,

\sqrt{x^8} = x^{8/2} = x^4

, and

\sqrt{y^6} = y^{6/2} = y^3

.

3

Divide the numerator by the denominator.

\frac{16x^6 y^4}{4x^4 y^3} = 4x^2 y^1

Divide the coefficients and subtract the exponents of the corresponding variables:

\frac{16}{4} = 4

,

x^{6-4} = x^2

, and

y^{4-3} = y^1

.

4

Identify the values of

a

,

b

, and

c

, and sum them.

a = 4

,

b = 2

,

c = 1

, and

a + b + c = 7

Match the simplified form

4x^2 y^1

to

ax^b y^c

to find

a

,

b

, and

c

, then calculate their sum.

Anahtar Kavram

Simplifying equivalent algebraic expressions using exponent rules and radical properties.

Tahmini Süre:1m 30s

Soru 58Soru →

For all positive values of $x$ , which of the following is equivalent to the expression $3x^{\frac{1}{2}} (2x^{\frac{3}{2}} - x^{-\frac{1}{2}})$ ?

6x^{\frac{3}{4}} - 3

6x^2 - 3x

6x^2 - 3

5x^2 - 3

Cevabı ve açıklamayı göster

Cevap: $6x^2 - 3$

Cevap

6x^2 - 3

Distributing the term

3x^{\frac{1}{2}}

to both terms inside the parentheses yields

3x^{\frac{1}{2}}(2x^{\frac{3}{2}}) - 3x^{\frac{1}{2}}(x^{-\frac{1}{2}})

. Multiplying the coefficients and adding the exponents according to the rule

x^a \cdot x^b = x^{a+b}

gives

6x^{\frac{1}{2} + \frac{3}{2}} - 3x^{\frac{1}{2} - \frac{1}{2}} = 6x^2 - 3x^0

. Since

x^0 = 1

for any positive

x

, the simplified equivalent expression is

6x^2 - 3

.

Adım Adım Çözüm

1

Distribute the term

3x^{\frac{1}{2}}

to both terms inside the parentheses.

3x^{\frac{1}{2}} \cdot 2x^{\frac{3}{2}} - 3x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}}

Apply the distributive property

a(b - c) = ab - ac

to expand the expression.

2

Multiply the coefficients and apply the product rule for exponents,

x^a \cdot x^b = x^{a+b}

, to each product.

(3 \cdot 2)x^{\frac{1}{2} + \frac{3}{2}} - 3x^{\frac{1}{2} + (-\frac{1}{2})}

When multiplying terms with the same base, keep the base and add the exponents.

3

Simplify the arithmetic in the exponents and evaluate the resulting terms.

6x^2 - 3

Since

\frac{1}{2} + \frac{3}{2} = 2

and

\frac{1}{2} - \frac{1}{2} = 0

, the expression simplifies to

6x^2 - 3x^0

. Because

x > 0

,

x^0 = 1

, making the final expression

6x^2 - 3

.

Anahtar Kavram

Equivalent Algebraic Expressions

Alternatif Yöntem

Substitute a simple value for

x

, such as

x = 4

. The original expression evaluates to

3(4)^{1/2}(2(4)^{3/2} - 4^{-\frac{1}{2}}) = 3(2)(2(8) - \frac{1}{2}) = 6(16 - 0.5) = 6(15.5) = 93

. Evaluating the correct expression

6x^2 - 3

at

x = 4

yields

6(16) - 3 = 96 - 3 = 93

. Evaluating the other options at

x = 4

yields different values.

Tahmini Süre:1m 30s

Soru 59Soru →

For all $x > 0$ , which of the following is equivalent to the expression $\frac{4x^2 - 25}{2x^{\frac{3}{2}} + 5x^{\frac{1}{2}}}$ ?

2x - 5

2x^{\frac{3}{2}} - 5x^{\frac{1}{2}}

2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}

2x^{\frac{1}{2}} + 5x^{-\frac{1}{2}}

Cevabı ve açıklamayı göster

Cevap: $2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}$

Cevap

The correct equivalent expression is

2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}

.

To simplify the expression

\frac{4x^2 - 25}{2x^{\frac{3}{2}} + 5x^{\frac{1}{2}}}

, we first factor the numerator and the denominator. The numerator is a difference of squares:

4x^2 - 25 = (2x - 5)(2x + 5)

. In the denominator, we can factor out

x^{\frac{1}{2}}

to get

x^{\frac{1}{2}}(2x + 5)

. Substituting these factored forms gives

\frac{(2x - 5)(2x + 5)}{x^{\frac{1}{2}}(2x + 5)}

. Canceling the common factor

(2x + 5)

yields

\frac{2x - 5}{x^{\frac{1}{2}}}

. Dividing each term in the numerator by the denominator gives

\frac{2x}{x^{\frac{1}{2}}} - \frac{5}{x^{\frac{1}{2}}}

. Applying exponent rules, this simplifies to

2x^{1 - \frac{1}{2}} - 5x^{-\frac{1}{2}} = 2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}

, which is the correct expression.

Adım Adım Çözüm

1

Factor the numerator of the expression as a difference of squares.

4x^2 - 25 = (2x - 5)(2x + 5)

To simplify the rational expression, we need to factor both the numerator and the denominator to identify common factors.

2

Factor out the common term

x^{\frac{1}{2}}

from the denominator.

2x^{\frac{3}{2}} + 5x^{\frac{1}{2}} = x^{\frac{1}{2}}(2x + 5)

Factoring out

x^{\frac{1}{2}}

reveals the common binomial factor

(2x + 5)

in the denominator.

3

Substitute the factored forms back into the original expression and cancel the common factor

(2x + 5)

.

\frac{(2x - 5)(2x + 5)}{x^{\frac{1}{2}}(2x + 5)} = \frac{2x - 5}{x^{\frac{1}{2}}}

Since

x > 0

, the term

2x + 5

is non-zero, allowing us to cancel it from both the numerator and the denominator.

4

Divide each term in the numerator by

x^{\frac{1}{2}}

and apply the rules of exponents.

\frac{2x}{x^{\frac{1}{2}}} - \frac{5}{x^{\frac{1}{2}}} = 2x^{1 - \frac{1}{2}} - 5x^{-\frac{1}{2}} = 2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}}

This simplifies the rational expression to its final equivalent form.

Anahtar Kavram

Equivalent Algebraic Expressions

Soru 60Soru →

For all $x \neq 2$ , the expression $\frac{3x^2 + kx - 8}{x - 2}$ is equivalent to $3x + 10 + \frac{12}{x - 2}$ , where $k$ is a constant. What is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of the constant

k

is

4

.

The correct answer is

4

. By rewriting the right-hand side of the equation with a common denominator of

x - 2

, the expression becomes

\frac{(3x + 10)(x - 2) + 12}{x - 2}

. Expanding and simplifying the numerator yields

3x^2 + 4x - 8

. Comparing this to the numerator of the left-hand side,

3x^2 + kx - 8

, shows that the coefficient of the

x

term,

k

, must equal

4

.

Adım Adım Çözüm

1

Set up the equation representing the equivalence of the two expressions.

\frac{3x^2 + kx - 8}{x - 2} = 3x + 10 + \frac{12}{x - 2}

This establishes the relationship that must hold for all

x \neq 2

.

2

Find a common denominator for the terms on the right-hand side.

3x + 10 + \frac{12}{x - 2} = \frac{(3x + 10)(x - 2)}{x - 2} + \frac{12}{x - 2}

To combine the terms into a single rational expression, they must have the same denominator.

3

Expand and simplify the numerator on the right-hand side.

(3x + 10)(x - 2) + 12 = 3x^2 + 4x - 8

Expanding the product of binomials and combining like terms yields the simplified numerator.

4

Equate the numerators of the simplified expressions to solve for

k

.

3x^2 + kx - 8 = 3x^2 + 4x - 8

, which means

k = 4

.

Since the denominators are equal, the numerators must be identical for all values of

x \neq 2

, meaning their corresponding coefficients must match.

Anahtar Kavram

Equivalent Algebraic Expressions

Alternatif Yöntem

An alternative method is to substitute a convenient value for

x

that is not equal to

2

. Substituting

x = 1

into both sides of the equivalence gives:

\frac{3(1)^2 + k(1) - 8}{1 - 2} = 3(1) + 10 + \frac{12}{1 - 2}

. Simplifying this yields

\frac{k - 5}{-1} = 13 - 12

, which simplifies to

-k + 5 = 1

. Solving for

k

results in

k = 4

.

Tahmini Süre:1m 30s

Equivalent Algebraic Expressions

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Alternatif Yöntem

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