ĐKXĐ: \(x\ne0\)
Ta có: \(2\left(x^2+\dfrac{1}{x^2}\right)-x-\dfrac{1}{x}-6=0\)
\(\Leftrightarrow2\left(x^2+\dfrac{1}{x^2}+2-2\right)-x-\dfrac{1}{x}-6=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2-\left(x+\dfrac{1}{x}\right)-10=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+4\left(x+\dfrac{1}{x}\right)-5\left(x+\dfrac{1}{x}\right)-10=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}+2\right)-5\left(x+\dfrac{1}{x}+2\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}+2\right)\left(2x+\dfrac{2}{x}-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}+2=0\\2x+\dfrac{2}{x}-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2}{x}+\dfrac{1}{x}+\dfrac{2x}{x}=0\\\dfrac{2x^2+2-5x}{x}=0\end{matrix}\right.\)
Suy ra: \(\left[{}\begin{matrix}x^2+2x+1=0\\2x^2-5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\\left(x-2\right)\left(2x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=2\left(nhận\right)\\x=\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-1;2;\dfrac{1}{2}\right\}\)
2(\(x^2+\dfrac{1}{x^2}\)) - (x + \(\dfrac{1}{x}\)) - 6 = 0
đặt x + \(\dfrac{1}{x}\) = a
<=>2(a2 - 2) - a - 6 = 0
<=> 2a2 - a - 10 = 0
<=> (2a - 5)(a + 2) = 0
TH1: a = \(\dfrac{5}{2}\)
=> x + \(\dfrac{1}{x}\) = \(\dfrac{5}{2}\)
<=> x2 + 1 = \(\dfrac{5x}{2}\)
<=> 2x2 - 5x + 1 = 0
<=> x1 = \(\dfrac{5+\sqrt{17}}{4}\) ;x2 = \(\dfrac{5-\sqrt{17}}{4}\)
TH2: a = -2
=> x + \(\dfrac{1}{x}\) = -2
<=> x2 + 2x + 1 = 0
<=> (x + 1)2 = 0
<=> x + 1 = 0
<=> x = -1
S = {\(\dfrac{5+\sqrt{17}}{4}\);\(\dfrac{5-\sqrt{17}}{4}\);-1}