`(x-3)/(x-2)+(x-2)/(x-4)=-1`         (ĐK:`x \ne 2,x \ne 4`)

`⇔((x-3)(x-4)+(x-2)(x-2))/((x-2)(x-4))=(-(x-2)(x-4))/((x-2)(x-4))`

`⇒x^2-4x-3x+12+x^2-4x+4=-x^2+4x+2x-8`

`⇔x^2+x^2+x^2-4x-3x-4x-4x-2x+12+4+8=0`

`⇔3x^2-17x+24=0`

`⇔3x^2-9x-8x+24=0`

`⇔3x(x-3)-8(x-3)=0`

`⇔(x-3)(3x-8)=0`

`⇔`\(\left[ \begin{array}{l}x-3=0\\3x-8=0\end{array} \right.\)

`⇔`\(\left[ \begin{array}{l}x=3\\x=\frac{8}{3}\end{array} \right.\) `(\text{t/m đk})`

Vậy ptr đã cho có tập nghiệm `S={8/3;3}`

Đáp án+Giải thích các bước giải:

`(x-3)/(x-2)+(x-2)/(x-4)=-1``ĐK:x≠2;x≠4`

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

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

`<=>x^2-7x+12+x^2-4x+4=-x^2+6x-8`

`<=>3x^2-17x+24=0`

`<=>3x^2-9x-8x+24=0`

`<=>3x(x-3)-8(x-3)=9`

`<=>(3x-8)(x-3)=0`

`<=>`\(\left[ \begin{array}{l}3x-8=0\\x-3=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{8}{3}(t/m)\\x=3(t/m)\end{array} \right.\) 

Vậy `S={8/3;3}`