Question

Simplify: $[\frac{1}{1+a}+\frac{2a}{1-a^2}]\times \frac{(a^2+4a-5)}{a^2+10a+25}$

• A. $\frac{-1}{a+1}$
• B. $\frac{1}{1-a}$
• C. $\frac{1}{a+5}$
• D. $\frac{-1}{a+5}$

Answer 1

The correct option is D $\frac{-1}{a+5}$

$(\frac{1}{1+a}+\frac{2a}{1-a^2})\times \frac{(a^2+4a-5)}{a^2+10a+25}$
= $\left(\frac{1-a+2a}{(1-a^2)}\right)\times \frac{(a+5)(a-1)}{{(a+5)}^2}$
= $\frac{1+a}{1-a^2}\times \frac{(a+5)(a-1)}{{(a+5)}^2}$
= $\frac{(1+a)}{(1+a)(1-a)}\times \frac{-(a+5)(1-a)}{{(a+5)}^2}=\frac{-1}{a+5}$