Question

Answer 1

Lời giải:

$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{(2x-1)(2x+1)}=\frac{49}{99}$\

$\Rightarrow \frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{(2x-1)(2x+1)}=\frac{98}{99}$

$\Rightarrow \frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{(2x+1)-(2x-1)}{(2x-1)(2x+1)}=\frac{98}{99}$

$\Rightarrow 1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}$

$\Rightarrow 1-\frac{1}{2x+1}=\frac{98}{99}$

$\Rightarrow \frac{1]{2x+1}=\frac{1}{99}$

$\Rightarrow 2x+1=99$

$\Rightarrow x=49$