\(\left(x+\dfrac{1}{x}\right)^2+\left(x+\dfrac{1}{x}\right)-8=0\)
Đặt \(x+\dfrac{1}{x}=t\)
\(t^2+t-8=0\Leftrightarrow t^2+t+\dfrac{1}{4}-\dfrac{1}{4}-8=0\)
\(\Leftrightarrow\left(t+\dfrac{1}{2}\right)^2-\dfrac{33}{4}=0\)
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