Sửa đề : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{8}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{8}\)
\(1-\frac{1}{x+1}=\frac{1}{8}\)
\(\frac{1}{x+1}=\frac{7}{8}\Leftrightarrow8=7x+7\Leftrightarrow x=\frac{1}{7}\)
(1/1.2 +1/2.3 1/3.4 + ....+1/8.9) x=23/45
(1/1.2 +1/2.3 1/3.4 + ....+1/8.9) x=23/45
1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3
a, 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/ x. (x + 1) = 201
1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=2015/2016
1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=19/20
(1/1.2+1/2.3+1/3.4+…+1/49.50)x=49/50
Tìm x, biết
1/1.2+1/2.3+1/3.4+...........+1/x.(x+1)= 2017/2018
($\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + ... + $\frac{1}{2011. 2012}$ ) x = 2011
