|x+1/1.2|+|x+1/2.3|+...+|x+1/99.100|=100x

                  Bài giải :

`|x+1/1xx2|+|x+1/2xx3|+...+|x+1/99xx100|=100x.`

Mà do vế trái luôn `: x > 0`

`=>`

`x + 1/1xx2+ x + 2/3 +x+3/4+.....+x+1/99xx100=100x.`

Vế trái có `99x`

`x = 1/1xx2 + 1/2xx3+1/3xx4+...+ 1/99xx100.`

`x = 1- 1/2 + 1/2 - 1/3 + 1/3 -1/4 +...+ 1/99-1/100.`

`x = 1 - 1/100.`

`x = 99/100.`

$|x+\dfrac{1}{1.2}| +|x+\dfrac{1}{2.3}|+...+|x+\dfrac{1}{99.100}|=100x(1)$

Do $\begin{cases} |x+\dfrac{1}{1.2}|\ge0\\|x+\dfrac{1}{2.3}|\ge0\\....\\|x+\dfrac{1}{99.10}|\ge0\end{cases}(∀x\in R)$

$\to |x+\dfrac{1}{1.2}| +|x+\dfrac{1}{2.3}|+...+|x+\dfrac{1}{99.100}|\ge 0 ∀x\in R$

Do đó : $100x\ge 0\to x\ge 0$

Khi đó $(1)$ có dạng :

$x+\dfrac{1}{1.2}+x+\dfrac{1}{2.3}+...+x+\dfrac{1}{99.100}=100x\\\to (x+x+...+x)+(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100})=100x\\\to 99x+(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100})=100x\\\to 99x + (1-\dfrac{1}{100})=100x\\\to 99x + \dfrac{99}{100}=100x\\\to -x=\dfrac{-99}{100}\\\to x=\dfrac{99}{100}(Tm)$
Vậy $x=\dfrac{99}{100}$