Question

Simplify: $\frac{a^{\frac{1}{2}}+a^{-\frac{1}{2}}}{1-a}+\frac{1-a^{-\frac{1}{2}}}{1-\sqrt{a}}$

• A. $1$
• B. $0$
• C. $\frac{2}{1-a}$
• D. $1+a$

Answer 1

Answer: (C)

$\frac{2}{1-a}$

$\frac{a^{\frac{1}{2}}+a^{\frac{-1}{2}}}{1-a}+\frac{1-a^{\frac{-1}{2}}}{1+\sqrt{a}}=\frac{\sqrt{a}+\frac{1}{\sqrt{a}}}{1-a}+\frac{1-\frac{1}{\sqrt{a}}}{1+\sqrt{a}}$

$=\frac{\frac{a+1}{\sqrt{a}}}{1-a}+\frac{\frac{\sqrt{a}-1}{\sqrt{a}}}{1+\sqrt{a}}$

$=\frac{a+1}{\sqrt{a}(1-a)}+\frac{\sqrt{a}-1}{\sqrt{a}(1+\sqrt{a})}$

$=\frac{(a+1)(1+\sqrt{a})+(\sqrt{a}-1)(1-a)}{\sqrt{a}(1-a)(1+\sqrt{a})}$

$=\frac{a+1+a\sqrt{a}+\sqrt{a}+\sqrt{a}-1-a\sqrt{a}+a}{\sqrt{a}(1-a)(1+\sqrt{a})}$

$=\frac{2a+2\sqrt{a}}{\sqrt{a}(1-a)(1+\sqrt{a})}$

$=\frac{2\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}(1-a)(1+\sqrt{a})}=\frac{2}{1-a}$