Đáp án:
`S ={1+\sqrt{3};1-\sqrt{3}}`
Giải thích các bước giải:
`(x-1)/(x-2) + (x+3)/(x-4) = 2/((x-2)(x-4)) `(dkxđ: $x\neq2 ; x\neq4)$
`⇔((x-1)(x-4))/((x-4)(x-2)) + ((x+3)(x-2))/((x-2)(x-4))= 2/((x-2)(x-4))`
`⇒(x-1)(x-4) + (x+3)(x-2)=2`
`⇔x^2-4x-x+4+x^2-2x+3x-6-2=0`
`⇔2x^2-4x - 4 =0`
`⇔2(x^2-2x-2)=0 `
`⇔x^2-2x-2 =0`
`⇔x^2-2x+1 =3`
`⇔(x-1)^2=3`
⇔ \(\left[ \begin{array}{l}x=1+\sqrt{3}(tm)\\x=1-\sqrt{3}(tm)\end{array} \right.\)
Vậy `S ={1+\sqrt{3};1-\sqrt{3}}`
