For all positive real numbers x and y, consider the algebraic expression: \frac{\left((x +… — ACT

For all positive real numbers $x$ and $y$ , consider the algebraic expression:

\frac{\left((x + y)y^2\right)^{1/2} (x + y)^{2/3}}{(x + y)^{-1/2}y}

Which of the following is equivalent to this expression?

A
$(x + y)^{5/6}$
B
$x^{5/3} + y^{5/3}$
$(x + y)^{5/3}$ Cevap
D
$(x + y)^{11/10}$
E
$(x + y)^{5/3} y$

Cevap

(x + y)^{5/3}

The correct answer is obtained by applying the properties of exponents step-by-step. First, apply the power of a product rule to the term

((x+y)y^2)^{1/2}

to get

(x+y)^{1/2}y

. Next, combine the terms in the numerator using the product rule of exponents to get

(x+y)^{7/6}y

. Finally, divide by the denominator using the quotient rule of exponents, which cancels the

y

terms and results in

(x+y)^{5/3}

Adım Adım Çözüm

Apply the power of a product property to simplify the first term in the numerator.

\left((x + y)y^2\right)^{1/2} = (x + y)^{1/2}(y^2)^{1/2} = (x + y)^{1/2}y

To distribute the exponent

1/2

to each factor inside the parentheses.

Multiply the simplified term by the other factor in the numerator.

(x + y)^{1/2}y \cdot (x + y)^{2/3} = (x + y)^{1/2 + 2/3}y = (x + y)^{7/6}y

To combine bases of the same value by adding their exponents:

\frac{1}{2} + \frac{2}{3} = \frac{3}{6} + \frac{4}{6} = \frac{7}{6}

Divide the simplified numerator by the denominator.

\frac{(x + y)^{7/6}y}{(x + y)^{-1/2}y} = (x + y)^{7/6 - (-1/2)} = (x + y)^{7/6 + 3/6} = (x + y)^{10/6} = (x + y)^{5/3}

To subtract the exponent in the denominator from the exponent in the numerator and cancel the common factor

y

Anahtar Kavram

Properties of Exponents in Algebraic Expressions

Alternatif Yöntem

Instead of simplifying the numerator first, we can divide the terms inside the parentheses first if we rewrite the expression, but the standard method of simplifying the numerator and then performing division is the most direct.

Tahmini Süre:1m 30s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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Alternatif Yöntem

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