Có \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)+x=\frac{23}{45}\)
Cho \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\)
Ta có công thức sau: \(\frac{1}{n\cdot\left(n+1\right)}+\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}=\frac{2}{n\cdot\left(n+1\right)\left(n+1\right)}\)
\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{9\cdot10}=\frac{22}{45}\)
\(\Rightarrow A=\frac{22}{45}:2=\frac{11}{45}\)
Thay vào phép tính trên ta được:
\(\frac{11}{45}\cdot x=\frac{23}{45}\\ x=\frac{23}{45}:\frac{11}{45}\\ x=\frac{23}{11}\)
Vậy \(x=\frac{23}{11}\)
