\(x^2+2x\sqrt{x+\dfrac{1}{x}}=8x-1\left(x\ne0\right)\)
Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge\dfrac{1}{8}\)
Vì \(x\ne0\Rightarrow\) chia 2 vế cho x,ta được:
\(x+2\sqrt{x+\dfrac{1}{x}}=8-\dfrac{1}{x}\Rightarrow x+\dfrac{1}{x}+2\sqrt{x+\dfrac{1}{x}}=8\)
Đặt \(\sqrt{x+\dfrac{1}{x}}=a\left(a>0\right)\)
pt trở thành \(a^2+2a-8=0\Rightarrow a^2-2a+4a-8=0\)
\(\Rightarrow a\left(a-2\right)+4\left(a-2\right)=0\Rightarrow\left(a-2\right)\left(a+4\right)=0\)
mà \(a>0\Rightarrow a=2\Rightarrow\sqrt{x+\dfrac{1}{x}}=2\Rightarrow x+\dfrac{1}{x}=4\)
\(\Rightarrow\dfrac{x^2-4x+1}{x}=0\Rightarrow x^2-4x+1=0\)
\(\Delta=\left(-4\right)^2-4=12\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-\sqrt{12}}{2}=2-\sqrt{3}\\x-\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+\sqrt{12}}{2}=2+\sqrt{3}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{2-\sqrt{3};2+\sqrt{3}\right\}\)
