For x > 1, which of the following expressions is equivalent to \frac{x - x^{-1}}{x^{1/2} -… — SAT

For $x > 1$ , which of the following expressions is equivalent to $\frac{x - x^{-1}}{x^{1/2} - x^{-1/2}}$ ?

x^{1/2} + x^{-1/2}

The expression

x^{1/2} + x^{-1/2}

is correct because the numerator

x - x^{-1}

can be factored using the difference of squares identity,

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

, where

a = x^{1/2}

and

b = x^{-1/2}

. This yields

x - x^{-1} = (x^{1/2} - x^{-1/2})(x^{1/2} + x^{-1/2})

. Substituting this back into the original fraction and canceling the common factor of

x^{1/2} - x^{-1/2}

in both the numerator and denominator simplifies the expression directly to

x^{1/2} + x^{-1/2}

Recognize the numerator

x - x^{-1}

as a difference of squares in terms of the base variables

x^{1/2}

and

x^{-1/2}

Write

x - x^{-1}

(x^{1/2})^2 - (x^{-1/2})^2 = (x^{1/2} - x^{-1/2})(x^{1/2} + x^{-1/2})

Since the denominator is

x^{1/2} - x^{-1/2}

, factoring the numerator as a difference of squares allows us to identify a common factor that can be canceled.

Substitute the factored numerator back into the original expression and cancel the common factor of

x^{1/2} - x^{-1/2}

from the numerator and the denominator.

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

For all

x > 1

, the term

x^{1/2} - x^{-1/2}

is non-zero, so we can divide both the numerator and the denominator by this common term to simplify the expression.

Factoring algebraic expressions using the difference of squares identity with fractional exponents.

Convert the fractional and negative exponents into standard algebraic fractions:

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

. Find common denominators for both the numerator and denominator to write the expression as

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

. Next, multiply the numerator by the reciprocal of the denominator:

\frac{(x - 1)(x + 1)}{x} \cdot \frac{\sqrt{x}}{x - 1}

. Cancel the common factor of

x - 1

to get

\frac{(x + 1)\sqrt{x}}{x}

. Distributing

\sqrt{x}

and dividing each term by

x

gives

\frac{x\sqrt{x}}{x} + \frac{\sqrt{x}}{x} = \sqrt{x} + \frac{1}{\sqrt{x}}

, which is equivalent to

x^{1/2} + x^{-1/2}

Tahmini Süre:1m 30s

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