Question

Simplify $\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+2)(x+3)}$

• A. $\frac{1}{(x+1)(x+3)}$
• B. $\frac{2}{(x+1)(x+3)}$
• C. $\frac{3}{(x+2)(x+3)}$
• D. $\frac{3}{(x+1)(x+3)}$

Answer 1

The correct option is D $\frac{3}{(x+1)(x+3)}$

$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+1)}=\frac{x+3+x+1+x+2}{(x+1)(x+2)(x+3)}$

$=\frac{3x+6}{(x+1)(x+2)(x+3)}$

$=\frac{3(x+2)}{(x+1)(x+2)(x+3)}=\frac{3}{(x+1)(x+3)}$