Đáp án:

 `a)S={4/3}`

`b)S={0}`

Giải thích các bước giải:

 `a)`

`\frac{1}{2x-3}-\frac{3}{x.(2x-3)}=5/x`

`(đk:x\ne0;x\ne 3/2)`

`<=>\frac{x}{x.(2x-3)}-\frac{3}{x.(2x-3)}=\frac{5.(2x-3)}{x.(2x-3)}`

`=>x-3=5.(2x-3)`

`<=>10x-15=x-3`

`<=>10x-x=-3+15`

`<=>9x=12`

`<=>x=4/3(tmđk)`

Vậy `S={4/3}`

`b)`

`\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{(x-5).(8-x)}-1`

`(đk:x\ne5;x\ne8)`

`<=>\frac{6.(x-8)}{(x-5).(x-8)}+\frac{2.(x-5)}{(x-5).(x-8)}=\frac{-18}{(x-5).(x-8)}-\frac{(x-5).(x-8)}{(x-5).(x-8)}`

`=>6x-48+2x-10=-18-[x.(x-8)-5.(x-8)]`

`<=>-18-(x^{2}-8x-5x+40)=8x-58`

`<=>-18-x^{2}+13x-40-8x+58=0`

`<=>-x^{2}+5x=0`

`<=>-x.(x-5)=0`

`<=>-x=0` hoặc `x-5=0`

`<=>x=0(tmđk)` hoặc `x=5(ktmđk)`

Vậy `S={0}.`