Question

Tìm x biết 1/1 +1/3 + 1/6 + ... + 2/x.(x+1) = 1\(\frac{2008}{2010}\)

Answer 1

Lời giải:

$\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x(x+1)}=1\frac{2008}{2010}$

$\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x(x+1)}=\frac{2009}{1005}$

$2(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)})=\frac{2009}{1005}$

$2(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1})=\frac{2009}{1005}$

$2(1-\frac{1}{x+1})=\frac{2009}{1005}$

$\frac{2x}{x+1}=\frac{2009}{1005}$

$\Rightarrow 2009(x+1)=2010x$

$\Rightarrow x=2009$