`b)|x+1/1.3|+|x+1/3.5|+|x+1/5.7|+...+|x+1/97.99|=50x`

Vì `|x+1/1.3| >=0 AA x`

`|1+1/3.5| >=0 AA x`

`....`

`|x+1/97.99 |>0 AA x`

`=>|x+1/1.3|+|x+1/3.5|+|x+1/5.7|+...+|x+1/97.99| >=0 AA x`

`=>50x >=0`

`=>x >=0`

Khi đó

`|x+1/1.3|+|x+1/3.5|+|x+1/5.7|+...+|x+1/97.99|=50x`

`x+1/1.3+x+1/3.5+x+1/5.7+...+x+1/97.99=50x`

`49x+(1/1.3+1/3.5+1/5.7+...+1/97.99)=50x`

`50x-49x=1/2(2/1.3+2/3.5+2/5.7+...+2/97.99)`

`x=1/2(1/1-1/3+1/3-1/5+1/5-1/7+..+1/97-1/99)`

`x=1/2(1-1/99)`

`x=49/99`

Vậy `x=49/99`

`c)`

`|x+1/1.5|+|x+1/5.9|+|x+1/9.13|+...+|1+1/397.401|=101x`

Ta có :

`|x+1/1.5| >=0 AA x`

`|x+1/5.9| >=0 AA x`

`....`

`|1+1/397.401| >=0 AA x`

`=>|x+1/1.5|+|x+1/5.9|+|x+1/9.13|+...+|x+1/397.401| >=0 AA x`

`=>101x  >=0`

`=>x >=0`

Khi đó

`|x+1/1.5|+|x+1/5.9|+|x+1/9.13|+...+|x+1/397.401|=101x`

`x+1/1.5+x+1/5.9+x+1/9.13+...+x+1/397.401=101x`

`100x+(1/1.5+1/5.9+1/9.13+...+1/397.401)=101x`

`101x-100x=1/4(4/1.5+4/5.9+4/9.13+...+4/397.401)`

`x=1/4(1/1-1/5+1/5-1/9+..+1/397-1/401)`

`x=1/4(1-1/401)`

`x=1/4*400/401`

`x=100/401`

Vậy `x=100/401`