giúp gải chi tiết với ạaaaaaaaaaaaaaa

Bài `23:`

`a)`

ĐKXĐ : `x\ne 1;x\ne 2`

`4/(x-1)-5/(x-2)=-3`

`<=>(4(x-2))/((x-1)(x-2))-(5(x-1))/((x-1)(x-2))=(-3(x-1)(x-2))/((x-1)(x-2))`

`=>4(x-2)-5(x-1)=-3(x-1)(x-2)`

`<=>4x-8-5x+5=-3(x^2-2x-x+2)`

`<=>-x-3=-3x^2+6x+3x-6`

`<=>3x^2+(-x-6x-3x)+(-3+6)=0`

`<=>3x^2-10x+3=0`

`<=>3x^2-9x-x+3=0`

`<=>3x(x-3)-(x-3)=0`

`<=>(x-3)(3x-1)=0`

`<=>x-3=0` hoặc `3x-1=0`

`<=>x=3(tm)` hoặc `x=1/3(tm)`

Vậy `S={3;1/3}`

`b)`

ĐKXĐ : `x\ne 1/3;x\ne -1/3`

`12/(1-9x^2)=(1-3x)/(1+3x)-(1+3x)/(1-3x)`

`<=>12/((1-3x)(1+3x))=((1-3x)^2)/((1-3x)(1+3x))-((1+3x)^2)/((1-3x)(1+3x))`

`=>12=(1-3x)^2-(1+3x)^2`

`<=>12=(1-3x-1-3x)(1-3x+1+3x)`

`<=>-6x.2=12`

`<=>-12x=12`

`<=>x=-1(tm)`

Vậy `S={-1}`

`#owen`

Đáp án+Giải thích các bước giải:

 Bài `23:`

`a)`

`\frac{4}{x-1}-\frac{5}{x-2}=-3`

`(đk:x\ne1;x\ne2)`

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

`=>4x-8-5x+5=-3.(x^{2}-2x-x+2)`

`<=>-x-3=-3.(x^{2}-3x+2)`

`<=>-3x^{2}+9x-6+x+3=0`

`<=>-3x^{2}+10x-3=0`

`<=>-3x^{2}+9x+x-3=0`

`<=>-3x.(x-3)+1.(x-3)=0`

`<=>-(3x-1).(x-3)=0`

`<=>(3x-1).(x-3)=0`

`<=>3x-1=0` hoặc `x-3=0`

`<=>3x=1` hoặc `x=3`

`<=>x=1/3(tm)` hoặc `x=3(tm)`

Vậy `S={1/3;3}`

`b)`

`\frac{12}{1-9x^{2}}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}`

`(đk:x\ne\pm1/3)`

`<=>\frac{12}{(1-3x).(1+3x)}=\frac{(1-3x).(1-3x)}{(1-3x).(1+3x)}-\frac{(1+3x).(1+3x)}{(1-3x).(1+3x)}`

`=>12=(1-3x)^{2}-(1+3x)^{2}`

`<=>(1-3x-1-3x).(1-3x+1+3x)=12`

`<=>(-6x).2=12`

`<=>-12x=12`

`<=>x=-1(tmđk)`

Vậy `S={-1}`