Đáp án+Giải thích các bước giải:

 `e)`

`\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)`

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

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

`<=>-9x=-12`

`<=>x=4/3(tmđk)`

Vậy `S={4/3}.`

`f)`

`\frac{x^{3}-(x-1)^{3}}{(4x+3).(x-5)}=\frac{7x-1}{4x+3}-\frac{x}{x-5}`

`(đk:x\ne-3/4;x\ne5)`

`<=>\frac{(x-x+1).[x^{2}+x.(x-1)+(x-1)^{2}]}{(4x+3).(x-5)}=\frac{(7x-1).(x-5)}{(4x+3).(x-5)}-\frac{x.(4x+3)}{(4x+3).(x-5)}`

`=>1.(x^{2}+x^{2}-x+x^{2}-2x+1)=7x^{2}-35x-x+5-4x^{2}-3x`

`<=>3x^{2}-3x+1=3x^{2}-39x+5`

`<=>3x^{2}-3x^{2}-3x+39x=5-1`

`<=>36x=4`

`<=>x=1/9(tmđk)`

Vậy `S={1/9}.`

`g)`

`\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{(x-1).(x+3)}`

`(đk:x\ne1;x\ne-3)`

`<=>\frac{(3x-1).(x+3)}{(x-1).(x+3)}-\frac{(2x+5).(x-1)}{(x-1).(x+3)}=\frac{(x-1).(x+3)}{(x-1).(x+3)}-\frac{4}{(x-1).(x+3)}`

`=>3x.(x+3)-1.(x+3)-(2x^{2}-2x+5x-5)=x^{2}+3x-x-3-4`

`<=>3x^{2}+9x-x-3-2x^{2}-3x+5=x^{2}+2x-7`

`<=>x^{2}+5x+2-x^{2}-2x+7=0`

`<=>3x+9=0`

`<=>3x=-9`

`<=>x=-3(ktmđk)`

`->` Phương trình vô nghiệm

Vậy phương trình vô nghiệm.

`h)`

`\frac{13}{(x-3).(2x+7)}+\frac{1}{2x+7}=\frac{6}{(x-3).(x+3)}`

`(đk:x\ne\pm3;x\ne -7/2)`

`<=>\frac{13.(x+3)}{(x-3).(x+3).(2x+7)}+\frac{(x-3).(x+3)}{(x-3).(x+3).(2x+7)}=\frac{6.(2x+7)}{(x-3).(x+3).(2x+7)}`

`=>13x+39+x^{2}-3^{2}=12x+42`

`<=>x^{2}+13x+30-12x-42=0`

`<=>x^{2}+x-12=0`

`<=>x^{2}+4x-3x-12=0`

`<=>x.(x+4)-3.(x+4)=0`

`<=>(x-3).(x+4)=0`

`<=>x-3=0` hoặc `x+4=0`

`<=>x=3(ktmđk)` hoặc `x=-4(tmđk)`

`<=>x=-4`

Vậy `S={-4}.`