Giải hộ mk từ câu 7 aj hứa sẽ vote 5sao mk đang cần gấp

Đáp án+Giải thích các bước giải:

 `7)`

`\frac{x+2}{x-2}-1/x=\frac{2}{x^{2}-2x}`

`(đk:x\ne0;x\ne2)`

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

`=>x^{2}+2x-(x-2)=2`

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

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

`<=>x.(x+1)=0`

`<=>x=0(ktmđk)` hoặc `x+1=0`

`<=>x=-1(tmđk)`

Vậy `S={-1}.`

`8)`

`\frac{2}{x+1}-\frac{3}{x+2}=\frac{1}{3x+3}`

`(đk:x\ne-1;x\ne-2)`

`<=>\frac{2.3.(x+2)}{3.(x+1).(x+2)}-\frac{3.3.(x+1)}{3.(x+1).(x+2)}=\frac{x+2}{3.(x+1).(x+2)}`

`=>6.(x+2)-9.(x+1)=x+2`

`<=>6x+12-9x-9=x+2`

`<=>-3x+3-x-2=0`

`<=>-4x+1=0`

`<=>-4x=-1`

`<=>x=1/4(tmđk)`

Vậy `S={1/4}`

`9)`

`\frac{3}{x+3}-\frac{1}{x-2}=\frac{5}{2x+6}`

`(đk:x\ne-3;x\ne2)`

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

`=>6.(x-2)-2.(x+3)=5x-10`

`<=>6x-12-2x-6-5x+10=0`

`<=>-x-8=0`

`<=>-x=8`

`<=>x=-8(tmđk)`

Vậy `S={-8}.`

`10)`

`\frac{6}{x+3}-\frac{1}{x-2}=\frac{5}{2x+6}`

`(đk:x\ne2;x\ne-3)`

`<=>\frac{6.2.(x-2)}{2.(x-2).(x+3)}-\frac{2.(x+3)}{2.(x-2).(x+3)}=\frac{5.(x-2)}{2.(x-2).(x+3)}`

`=>12.(x-2)-2.(x+3)=5x-10`

`<=>12x-24-2x-6=5x-10`

`<=>10x-30-5x+10=0`

`<=>5x-20=0`

`<=>5x=20`

`<=>x=4(tmđk)`

Vậy `S={4}.`

`11)`

`\frac{7}{2x-3}+\frac{1}{2x-2}=\frac{3}{x-1}`

`(đk:x\ne 3/2;x\ne 1)`

`<=>\frac{7.2.(x-1)}{2.(x-1).(2x-3)}+\frac{2x-3}{2.(x-1).(2x-3)}=\frac{2.(2x-3).3}{2.(x-1).(2x-3)}`

`=>14.(x-1)+2x-3=6.(2x-3)`

`<=>14x-14+2x-3=12x-18`

`<=>16x-17-12x+18=0`

`<=>4x+1=0`

`<=>4x=-1`

`<=>x=-1/4(tmđk)`

Vậy `S={-1/4}.`

`12)`

`\frac{1}{2x-1}+\frac{3}{12x-8}=\frac{2}{3x-2}`

`(đk:x\ne 1/2; x\ne 2/3)`

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

`=>12x-8+6x-3=8.(2x-1)`

`<=>18x-11=16x-8`

`<=>18x-16x=-8+11`

`<=>2x=3`

`<=>x=3/2(tmđk)`

Vậy `S={3/2}.`