ĐKXĐ: \(x^3-2>=0\)
=>\(x^3\ge2\)
=>\(x\ge\sqrt[3]{2}\)
\(\sqrt[3]{x^2-1}+x=\sqrt{x^3-2}\)
=>\(\sqrt[3]{x^2-1}-2+x-3=\sqrt{x^3-2}-5\)
=>\(\frac{x^2-1-8}{\sqrt[3]{\left(x^2-1\right)^2}+2\cdot\sqrt[3]{x^2-1}+4}+x-3=\frac{x^3-2-25}{\sqrt{x^3-2}+5}\)
=>\(\frac{x^2-9}{\sqrt[3]{\left(x^2-1\right)^2}+2\cdot\sqrt[3]{x^2-1}+4}+x-3=\frac{x^3-27}{\sqrt{x^3-2}+5}\)
=>\(\frac{\left(x-3\right)\left(x+3\right)}{\sqrt[3]{\left(x^2-1\right)^2}+2\cdot\sqrt[3]{x^2-1}+4}+x-3=\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\sqrt{x^3-2}+5}\)
=>\(\left(x-3\right)\left\lbrack\frac{\left(x+3\right)}{\sqrt[3]{\left(x^2-1\right)^2}+2\cdot\sqrt[3]{x^2-1}+4}+1\right\rbrack=\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\sqrt{x^3-2}+5}\)
=>\(\left(x-3\right)\left\lbrack\frac{\left(x+3\right)}{\sqrt[3]{\left(x^2-1\right)^2}+2\cdot\sqrt[3]{x^2-1}+4}+1\right\rbrack-\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\sqrt{x^3-2}+5}\) =0
=>\(\left(x-3\right)\left(\frac{\left(x+3\right)}{\sqrt[3]{\left(x^2-1\right)^2}+2\cdot\sqrt[3]{x^2-1}+4}+1-\frac{x^2+3x+9}{\sqrt{x^3-2}+5}\right)=0\)
=>x-3=0
=>x=3(nhận)
