Bài `3`

`a)`

`1/x+1/(2x)=3/2 (đk:x\ne0)`

`1.2+1=3x`

`3=3x`

`x=1 (tm)`

Vậy `S={1}`

`b)`

`(3x)/(4x-3)=-2 (đk:x\ne3/4)`

`-2(4x-3)=3x`

`-8x+6=3x`

`-8x-3x=-6`

`-11x=-6`

`x=6/11 (tm)`

Vậy `x=6/11`

`c)`

`2/(x-3)=1/(x+2) (đk:x\ne3;x\ne-2)`

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

`2x+4=x-3`

`2x-x=-3-4`

`x=-7 (tm)`

Vậy `S={-7}`

`d)`

`2/(x^2-2x)+1/x=(x+2)/(x-2) (đk:x\ne0;x\ne2)`
`2/(x(x-2))+1/x=(x+2)/(x-2)`

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

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

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

`x^2+x=0`

`x(x+1)=0`

`x=0 (ktm)` hoặc `x=-1 (tm)`

Vậy `S={-1}`

`e)`

`(3x-2)/(x+7)=(6x+1)/(2-3x) (đk:x\ne-7;x\ne2/3)`

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

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

`-9x^2+12x-4-6x^2-43x-7=0`

`-15x^2-31x-11=0`

`15x^2+31x+11=0`

`15(x^2+31/15x+11/15)=0`

`x^2+31/15x+11/15=0`

`x^2+31/15x+961/900-301/900=0`

`(x+31/30)^2=301/900`

`x+31/30=(\sqrt{301})/30` hoặc `x+31/30=-(\sqrt{301})/30`

`x=(\sqrt{301}-31)/30 (tm)` hoặc `x=(-\sqrt{301}-31)/30 (tm)`

Vậy `S={(\sqrt{301}-31)/30;(-\sqrt{301}-31)/30}`

`f)`

`(x+2)/(x-2)-5/x=8/(x^2-2x) (đk:x\ne0;x\ne2)`

`(x+2)/(x-2)-5/x=8/(x(x-2))`

`x(x+2)-5(x-2)=8`

`x^2+2x-5x+10=8`

`x^2-3x+10-8=0`

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

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

`x(x-1)-2(x-1)=0`

`(x-2)(x-1)=0`

`x=2 (ktm)` hoặc `x=1 (tm)`

Vậy `S={1}`

`g)`

`(x+2)/(x-2)-(x-2)/(x+2)=(-4)/(x^2-4) (đk:x\ne2;x\ne-2)`

`(x+2)/(x-2)-(x-2)/(x+2)=(-4)/((x-2)(x+2))`

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

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

`4.2x=-4`

`2x=-1`

`x=-1/2 (tm)`

Vậy `S={-1/2}`