Which choice is equivalent to the fraction below when [tex]\( x \geq 1 \)[/tex]?

Hint: Rationalize the denominator and simplify.

[tex]\[ \frac{1}{\sqrt{x}-\sqrt{x-1}} \][/tex]

A. [tex]\(-\sqrt{x}-\sqrt{x-1}\)[/tex]

B. [tex]\(\sqrt{x-1}-\sqrt{x}\)[/tex]

C. [tex]\(\frac{\sqrt{x}+\sqrt{x-1}}{2x-1}\)[/tex]

D. [tex]\(\sqrt{x}+\sqrt{x-1}\)[/tex]



Answer :

To solve the given problem, we need to rationalize the denominator of the fraction:

[tex]\[ \frac{1}{\sqrt{x}-\sqrt{x-1}} \][/tex]

Rationalizing the denominator involves multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of [tex]\( \sqrt{x} - \sqrt{x-1} \)[/tex] is [tex]\( \sqrt{x} + \sqrt{x-1} \)[/tex]. Let's proceed with this process:

1. Multiply by the Conjugate:

[tex]\[ \frac{1}{\sqrt{x}-\sqrt{x-1}} \times \frac{\sqrt{x} + \sqrt{x-1}}{\sqrt{x} + \sqrt{x-1}} \][/tex]

2. Simplify the Numerator:

The numerator now becomes:

[tex]\[ 1 \times (\sqrt{x} + \sqrt{x-1}) = \sqrt{x} + \sqrt{x-1} \][/tex]

3. Simplify the Denominator:

The denominator now becomes:

[tex]\[ (\sqrt{x} - \sqrt{x-1}) \times (\sqrt{x} + \sqrt{x-1}) \][/tex]

Using the difference of squares formula, [tex]\(a^2 - b^2 = (a-b)(a+b)\)[/tex], we have:

[tex]\[ (\sqrt{x})^2 - (\sqrt{x-1})^2 \][/tex]

Simplifying further:

[tex]\[ x - (x-1) = x - x + 1 = 1 \][/tex]

4. Combine the Results:

After simplifying both the numerator and the denominator, the fraction becomes:

[tex]\[ \frac{\sqrt{x} + \sqrt{x-1}}{1} \][/tex]

So the simplified form of the given fraction is:

[tex]\[ \sqrt{x} + \sqrt{x-1} \][/tex]

Thus, the equivalent choice is:

D. [tex]\( \sqrt{x} + \sqrt{x-1} \)[/tex]