Question

$\frac{y}{y-2}+\frac{1}{y+2}+\frac{3y}{4-y^2}$ = .

• A. $\frac{y^2-2}{y^2-4}$
• B. $\frac{y^2-2}{y^2+2}$
• C. $\frac{y+2}{y-2}$
• D. $\frac{1}{y+2}$

Answer 1

The correct option is A $\frac{y^2-2}{y^2-4}$

$\frac{y}{y-2}+\frac{1}{y+2}+\frac{3y}{4-y^2}$
= $\left[\frac{y(y+2)+1(y-2)}{(y-2)(y+2)}\right]+\frac{3y}{4-y^2}$
= $\frac{y^2+3y-2}{y^2-4}-\frac{3y}{y^2-4}$
= $\frac{y^2-2}{y^2-4}$