Bài `1`

`(3x-2)/(x+7)=(6x+1)/(2x-3) (đk:x ne -7;x ne 3/2)`

`(3x-2)(2x-3)=(6x+1)(x+7)`

`6x^2-9x-4x+6=6x^2+42x+x+7`

`6x^2-13x+6=6x^2+43x+7`

`6x^2-13x-6x^2-43x=7-6`

`-56x=1`

`x=-1/56` (tm)

Vậy `S={-1/56}`

Bài `2`

`(x+1)/(x-1)-(x-1)/(x+1)=4/(x^2-1) (đk: x ne 1;x ne -1)`

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

`(x+1)^2-(x-1)^2=4`

`(x+1-x+1)(x+1+x-1)=4`

`2.2x=4`

`4x=4`

`x=1` (ktm)

Vậy pt vô nghiệm

Bài `3`

`(x^2-x)/(x+3)-(x^2)/(x-3)=(7x^2-3x)/(9-x^2) (đk: x ne 3;x ne -3)`

`(x^2-x)/(x+3)-(x^2)/(x-3)=(3x-7x^2)/((x-3)(x+3))`

`(x^2-x)(x-3)-x^2(x+3)=3x-7x^2`

`x^3-3x^2-x^2+3x-x^3-3x^2=3x-7x^2`

`-7x^2+3x=3x-7x^2`

`-7x^2+3x-3x+7x^2=0`

`0x=0` (luôn đúng)

Vậy pt có vô số nghiệm

Bài `4` (đúng đề)

`1/(x-1)-(3x^2)/(x^2-1)=(2x)/(x^2+x+1) (đk: xne 1;x ne -1)`

`1/(x-1)-(3x^2)/((x-1)(x+1))=(2x)/(x^2+x+1)`

`(x+1)(x^2+x+1)-3x^2(x^2+x+1)=2x(x-1)(x+1)`

`x^3+x^2+x+x^2+x+1-3x^4-3x^3-3x^2=2x(x^2-1)`

`-3x^4-2x^3-x^2+2x+1=2x^3-2x`

`3x^4+2x^3+x^2-2x-1+2x^3-2x=0`

`3x^4+4x^3+x^2-4x-1=0`

Bài `4` (Nếu là `x^3-1)`

`1/(x-1)-(3x^2)/(x^3-1)=(2x)/(x^2+x+1) (đk: xne 1;x ne -1)`

`1/(x-1)-(3x^2)/((x-1)(x^2+x+1))=(2x)/(x^2+x+1)`

`x^2+x+1-3x^2=2x(x-1)`

`-2x^2+x+1=2x^2-2x`

`2x^2-2x+2x^2-x-1=0`

`4x^2-3x-1=0`

`4x^2-4x+x-1=0`

`4x(x-1)+(x-1)=0`

`(4x+1)(x-1)=0`

`4x+1=0` hoặc `x-1=0`

`x=-1/4` (tm) hoặc `x=1` (ktm)

Vậy `S={-1/4}`

Bài `5`

`3/(1-4x)=2/(4x+1)-(8+6x)/(16x^2-1) (đk: x ne 1/4;x ne -1/4)`

`3/(1-4x)=2/(1+4x)+(8+6x)/((1-4x)(1+4x))`

`3(1+4x)=2(1-4x)+8+6x`

`3+12x=2-8x+8+6x`

`12x+3=-2x+10`

`12x+2x=10-3`

`14x=7`

`x=1/2` (tm)

Vậy `S={1/2}`