Đáp án:
$S=\{2\}$
Giải thích các bước giải:
`ĐKXĐ: x\ne 1`
`x^2+(\frac{x}{x-1})^2=8`
`⇔\frac{x^2(x-1)^2}{(x-1)^2}+\frac{x^2}{(x-1)^2}=\frac{8(x-1)^2}{(x-1)^2}`
`⇒x^2(x^2-2x+1)+x^2=8(x^2-2x+1)`
`⇔x^4-2x^3+x^2+x^2=8x^2-16x+8`
`⇔x^4-2x^3+2x^2=8x^2-16x+8`
`⇔x^4-2x^3-6x^2+16x-8=0`
`⇔x^4+2x^3-2x^2-4x^3-8x^2+8x+4x^2+8x-8=0`
`⇔x^2(x^2+2x-2)-4x(x^2+2x-2)+4(x^2+2x-2)=0`
`⇔(x^2-4x+4)(x^2+2x-2)=0`
`⇔(x-2)^2(x^2+2x-2)=0`
\(⇔\left[ \begin{array}{l}(x-2)^2=0\\x^2+2x-2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2(tm)\\x^2+2x-2\end{array} \right.\)(vô nghiệm)
Vậy $S=\{2\}$
