ĐK: `x`$\neq$ `-1;-2;-3;-4;-5`

 $\frac{1}{(x+1)(x+2)}$ + $\frac{1}{(x+2)(x+3)}$ + $\frac{1}{(x+3)(x+4)}$ + $\frac{1}{(x+4)(x+5)}$ 

=$\frac{1}{x+1}$ -$\frac{1}{x+2}$ +$\frac{1}{x+2}$ -$\frac{1}{x+3}$ +$\frac{1}{x+3}$ -$\frac{1}{x+4}$ +$\frac{1}{x+4}$ - $\frac{1}{x+5}$ 

=$\frac{1}{x+1}$ - $\frac{1}{x+5}$ 

=$\frac{x+5-x-1}{(x+1)(x+5)}$ 

=$\frac{4}{(x+1)(x+5)}$