For all real numbers x > 0 such that x \neq 1, the algebraic expression \frac{\sqrt{x} - 1… — GMAT

For all real numbers $x > 0$ such that $x \neq 1$ , the algebraic expression $\frac{\sqrt{x} - 1}{\sqrt[4]{x} + 1} + \frac{\sqrt{x} + 1}{\sqrt[4]{x} - 1}$ is equivalent to $\frac{2(\sqrt[4]{x}^3 + 1)}{\sqrt{x} - 1}$ .

Cevap: Cevap

Cevap

The statement is True.

The statement is true because substituting

u = \sqrt[4]{x}

allows both expressions to be simplified via algebraic factoring identities to the identical expression

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

Adım Adım Çözüm

Perform a substitution to simplify the radical exponents.

Let

u = \sqrt[4]{x}

, so that

\sqrt{x} = u^2

. The left-hand side becomes

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

Converting fourth roots and square roots into polynomial terms makes factoring easier.

Simplify the left-hand side expression.

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

Factoring the numerator of the first fraction cancels out the

(u+1)

term prior to combining terms.

Factor the right-hand side expression using polynomial identities.

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

Applying the sum of cubes identity

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

allows cancellation of

(u+1)

from the denominator.

Compare the simplified left-hand side and right-hand side expressions.

Both expressions reduce to

\frac{2(\sqrt[4]{x}^2 - \sqrt[4]{x} + 1)}{\sqrt[4]{x} - 1}

Because both sides reduce to the identical simplified form for all

x > 0, x \neq 1

, the equivalence holds true.

Anahtar Kavram

Algebraic manipulation of radicals using fractional exponent substitution, difference of squares, and sum of cubes factoring.

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