\(2\left(x^2+\dfrac{1}{x^2}\right)+3\left(x+\dfrac{1}{x}\right)-16=0\left(1\right)\)

Condition: \(x\ne0\)

Put \(t=x+\dfrac{1}{x}\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2\)

\(\left(1\right)\Leftrightarrow2t^2+3t-20=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=-4\\t=\dfrac{5}{2}\end{matrix}\right.\)

If \(t=-4\Rightarrow x=-2\pm\sqrt{3}\)

If \(t=\dfrac{5}{2}\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

Conclude:...

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