Question

Simplify: $\left[\frac{{(x+y)}^{-1}.{(x-y)}^{-m}}{{(x+y)}^{m+n}.{(x-y)}^{1+n}}\right]{(x^2-y^2)}^{1+m+n}$
(where $x\ne y$ and $y\ne 0$)

• A. $x^2$
• B. x
• C. 0
• D. 1

Answer 1

The correct option is D 1

$\left[\frac{(x+y{)}^{-1}.(x-y{)}^{-m}}{(x+y{)}^{m+n}.(x-y{)}^{1+n}}\right]{(x^2-y^2)}^{1+m+n}$
$=\left[\right(x+y{)}^{-1-m-n}.(x-y{)}^{-m-1-n}]{(x^2-y^2)}^{1+m+n}$
$=\left[\right(x+y\left)\right(x-y){]}^{-1-m-n}{(x^2-y^2)}^{1+m+n}$
$=\left[\right(x^2-y^2{)}^{-1-m-n}]{(x^2-y^2)}^{1+m+n}$
$=(x^2-y^2{)}^{-1-m-n+1+m+n}=(x^2-y^2{)}^0=1$