\(\left|x+\frac{1}{1\cdot5}\right|+\left|x+\frac{1}{5\cdot9}\right|+...+\left|x+\frac{1}{397\cdot401}\right|=101x\left(1\right)\)
Điều kiện:\(101x\ge0\)\(\Rightarrow\left|x+\frac{1}{1\cdot5}\right|\ge0;\left|x+\frac{1}{5\cdot9}\right|\ge0;.....;\left|x+\frac{1}{397\cdot401}\right|\ge0\)
Do vậy\(\left(1\right)\)trở thành:\(x+\frac{1}{1\cdot5}+x+\frac{1}{5\cdot9}+...+x+\frac{1}{397\cdot401}=101x\)
\(\left(x+x+x+..+x\right)+\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+..+\frac{1}{397\cdot401}\right)\)
Có 100 số x
\(\Leftrightarrow\)\(100x+\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{397}-\frac{1}{401}\right)=101x\)
\(\Leftrightarrow\)\(100x+\frac{1}{4}\left(1-\frac{1}{401}\right)=101x\)
\(\Leftrightarrow100x+\frac{1}{4}\left(\frac{400}{401}\right)=101x\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\cdot\frac{400}{401}\)\(=\frac{100}{401}\)
