Giải phương trình: $\sqrt[3]{x^2-1}+x=\sqrt{x^3-2}$

Đáp án: $x=3$

Giải thích các bước giải:

$\sqrt[3]{{{x}^{2}}-1}+x=\sqrt{{{x}^{3}}-2}$   (ĐK: $x\ge \sqrt[3]{2}$)

$\Leftrightarrow \left[ \sqrt[3]{{{x}^{2}}-1}-\left( x-1 \right) \right]-\left[ \sqrt{{{x}^{3}}-2}-\left( 2x-1 \right) \right]=0$

$\Leftrightarrow \dfrac{\left( {{x}^{2}}-1 \right)-{{\left( x-1 \right)}^{3}}}{{{\left( \sqrt[3]{{{x}^{2}}-1} \right)}^{2}}+\left( x-1 \right)\sqrt[3]{{{x}^{2}}-1}+{{\left( x-1 \right)}^{2}}}-\dfrac{\left( {{x}^{3}}-2 \right)-{{\left( 2x-1 \right)}^{2}}}{\sqrt{{{x}^{3}}-2}+2x-1}=0$

$\Leftrightarrow \dfrac{-{{x}^{3}}+4{{x}^{2}}-3x}{{{\left( \sqrt[3]{{{x}^{2}}-1} \right)}^{2}}+\left( x-1 \right)\sqrt[3]{{{x}^{2}}-1}+{{\left( x-1 \right)}^{2}}}-\dfrac{{{x}^{3}}-4{{x}^{2}}+4x-3}{\sqrt{{{x}^{3}}-2}+2x-1}=0$

$\Leftrightarrow \dfrac{-x\left( x-1 \right)\left( x-3 \right)}{{{\left( \sqrt[3]{{{x}^{2}}-1} \right)}^{2}}+\left( x-1 \right)\sqrt[3]{{{x}^{2}}-1}+{{\left( x-1 \right)}^{2}}}-\dfrac{\left( x-3 \right)\left( {{x}^{2}}-x+1 \right)}{\sqrt{{{x}^{3}}-2}+2x-1}=0$

$\Leftrightarrow \left( x-3 \right)\left[ \dfrac{-x\left( x-1 \right)}{{{\left( \sqrt[3]{{{x}^{2}}-1} \right)}^{2}}+\left( x-1 \right)\sqrt[3]{{{x}^{2}}-1}+{{\left( x-1 \right)}^{2}}}-\dfrac{{{x}^{2}}-x+1}{\sqrt{{{x}^{3}}-2}+2x-1} \right]=0$

$\Leftrightarrow x=3$ (nhận) (vế sau luôn nhận giá trị âm với $x\ge \sqrt[3]{2}$)