ĐK: \(x\ge\sqrt[3]{2}\)
\(\Leftrightarrow\sqrt{x^3-2}-\sqrt[3]{x^2-1}-x=0\)
\(\Leftrightarrow\sqrt{x^3-2}-5-\sqrt[3]{x^2-1}+2-x+3=0\)
\(\Leftrightarrow\dfrac{x^3-27}{\sqrt{x^3-2}+5}-\dfrac{x^2-1-8}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1=0\end{matrix}\right.\)
Vì \(\dfrac{x^2+3x+9}{\sqrt{x^3-2}+5}-\dfrac{x+3}{\sqrt[3]{x^2-1}^2+2\sqrt[3]{x^2-1}+4}-1>0\forall x\ge\sqrt[3]{2}\)
nên x=3.
