Tìm x 1/3 + 1/6 +1/10 +...+2/x.(x+1)=2001/2003

Đáp án:

x=2002

Giải thích các bước giải:

ta có $\frac{2}{x(x+1)}=2.(\frac{1}{x}-\frac{1}{x+1})$

$=>\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{2001}{2003}$

$=>\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x(x+1)}=\frac{2001}{2003}$

$=>2.(\frac{1}{2}-\frac{1}{3})+2.(\frac{1}{3}-\frac{1}{4})+...+2.(\frac{1}{x}-\frac{1}{x+1})=\frac{2001}{2003}$

$=> 2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003})$

$=> 2(\frac{1}{2}-\frac{1}{x+1})=\frac{2001}{2003}$

$=>\frac{x-1}{x+1}=\frac{2001}{2003}$

$=> x=2002$

Đáp án: x=2002

Giải thích các bước giải:

VT= 2/6+2/12+2/20+....+2/ (x.(x+1))

=2/(2.3) +2/(3.4) +2/(4.5) +....+2/x.(x+1)

=2.[ 1/(2.3) +1/(3.4) +1/(4.5)+....+ 1/x.(x+1)]

=2.[ 1/2 - 1/3 + 1/3 -1/4 +1/4 - 1/5 +....+ 1/x - 1/(x+1)]

= 2. [ 1/2- 1/(x+1)]

= (x-1)/(x+1) = 2001/2003

vậy x= 2002