\(\Leftrightarrow x^2\left(2x^2-7x+10-\dfrac{7}{x}+\dfrac{2}{x^2}\right)=0\)
\(\Leftrightarrow2x^2-7x+10-\dfrac{7}{x}+\dfrac{2}{x^2}=0\)
\(\Leftrightarrow\left(2x^2+\dfrac{2}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+10=0\)
\(\Leftrightarrow2\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+10=0\)
Đặt \(x+\dfrac{1}{x}\) là a ,thì \(x^2+\dfrac{1}{x^2}\) là a2-2 ta được
2(a2-2)-7a+10=0
⇔2a2-4-7a+10=0
⇔2a2-7a+6=0
⇔2a2-4a-3a+6=0
⇔(2a2-4a)-(3a-6)=0
⇔2a(a-2)-3(a-2)=0
⇔(a-2)(2a-3)=0
\(\Rightarrow\left[{}\begin{matrix}a-2=0\\2a-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=2\\a=\dfrac{3}{2}\end{matrix}\right.\)
ta được a=2 => \(x+\dfrac{1}{x}=2\) => x=1
a=\(x+\dfrac{1}{x}=\dfrac{3}{2}\)(vô nghiệm)
vậy S={1}
