Question

Answer 1

Ta có: \(\left|x+\frac12\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|=120x\)

=>120x>=0

=>x>=0

Phương trình sẽ trở thành:

\(120x=x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+\cdots+x+\frac{1}{99\cdot100}\)

=>\(120x=99x+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{99\cdot100}\right)\)

=>\(21x=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

=>\(x=\frac{99}{100}:21=\frac{99}{2100}=\frac{33}{700}\) (nhận)