`|x+1/1.2|+|x+1/2.3|+.....+|x+1/149.150| =150x`

Ta có:

`|x+1/1.2| >=0`

`|x+1/2.3| >=0`

` ....`

`|x+1/149.150| >=0 `

Nên `150x>=0`

Suy  ra  `x>=0`

`|x+1/1.2|+|x+1/2.3|+.....+|x+1/149.150| =150x`

`x+1/1.2+x+1/2.3+.....+x+1/149.150 =150x`

`\underbrace{x+x+x .... + x}_{149 số x}+(1/1.2+1/2.3+....+1/149.150)=150x`  

`149x+[(1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + … + (1/149 − 1/150)]=150x`

`149x  +[1 − 1/150]=150x` 

`149x+ 149/150=150x`

`150x-149x=149/150`

`x=149/150`