$\frac{x+3}{x-2}$= $\frac{x-1}{x-4}$

Đáp án:

 `\frac{x+3}{x-2}=\frac{x-1}{x-4}` `(đk: x\ne2;x\ne4)`

`=>(x+3).(x-4)=(x-2).(x-1)`

`=>x.(x-4)+3.(x-4)=x.(x-1)-2.(x-1)`

`=>(x.x-x.4)+(3.x-3.4)=(x.x-x.1)-(2.x-2.1)`

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

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

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

`=>(x^{2}-x^{2})+(3x-x)-(12+2)=0`

`=>2x-14=0`

`=>2x=14`

`=>x=7(tmđk)`

Vậy `x=7` với `x\ne2;x\ne4`

ÁP DỤNG `axx(b+c)=axxb+axxc`