bài khó

V

vansang02121998

Tổng quát

$\frac{k}{a(a+k)}=\frac{1}{a}-\frac{1}{a+k}$




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

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

$=\frac{1}{x}$
 
Top Bottom