Đáp án:
x=2002
Giải thích các bước giải:
ta có $\frac{2}{x(x+1)}=2.(\frac{1}{x}-\frac{1}{x+1})$
$=>\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{2001}{2003}$
$=>\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x(x+1)}=\frac{2001}{2003}$
$=>2.(\frac{1}{2}-\frac{1}{3})+2.(\frac{1}{3}-\frac{1}{4})+...+2.(\frac{1}{x}-\frac{1}{x+1})=\frac{2001}{2003}$
$=> 2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003})$
$=> 2(\frac{1}{2}-\frac{1}{x+1})=\frac{2001}{2003}$
$=>\frac{x-1}{x+1}=\frac{2001}{2003}$
$=> x=2002$
