\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|=100x\)
=> \(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)
=> \(\left[x+x+x+...+x\right]+\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{99\cdot100}\right]=100x\)
=> \(99x+\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right]=100x\)
=> \(99x+\left[1-\frac{1}{100}\right]=100x\)
=> \(99x+\frac{99}{100}=100x\)
=> \(100x-99x=\frac{99}{100}\)
=> \(x=\frac{99}{100}\)
Check lại có đúng không nhé
Đúng 0
Bình luận (0)
