ĐK $x\ge \sqrt[3]{2}$. 

PT$\Leftrightarrow \left( \sqrt[3]{{{x}^{2}}-1}-2 \right)+\left( x-3 \right)=\left( \sqrt{{{x}^{3}}-2}-5 \right)$

$\Leftrightarrow \frac{{{x}^{2}}-9}{\sqrt[3]{{{\left( {{x}^{2}}-1 \right)}^{2}}}+2\sqrt[3]{{{x}^{2}}-1}+4}+\left( x-3 \right)=\frac{{{x}^{3}}-27}{\sqrt{{{x}^{3}}-2}+5}$

$\Leftrightarrow x=3$ hoặc $\frac{x+3}{\sqrt[3]{{{\left( {{x}^{2}}-1 \right)}^{2}}}+2\sqrt[3]{{{x}^{2}}-1}+4}+1=\frac{{{x}^{2}}+3x+9}{\sqrt{{{x}^{3}}-2}+5}(2)$

• $\sqrt[3]{{{\left( {{x}^{2}}-1 \right)}^{2}}}+2\sqrt[3]{{{x}^{2}}-1}+4>\sqrt[3]{{{\left( x-1 \right)}^{3}}}+2\sqrt[3]{{{x}^{2}}-1}+4>x+3$, suy ra VT(2) < 2.

• ${{x}^{2}}+3x+9=\left( {{x}^{2}}+x+1 \right)+\left( x-1 \right)+x+9\ge 2\sqrt{{{x}^{2}}-1}+x+9>2\left( \sqrt{{{x}^{2}}-2}+5 \right)$,

suy ra VP(2) > 2. Do đó, (2) vô nghiệm.

 

${{x}^{2}}+3x+9=\left( {{x}^{2}}+x+1 \right)+\left( x-1 \right)+x+9\ge 2\sqrt{{{x}^{2}}-1}+x+9>2\left( \sqrt{{{x}^{2}}-2}+5 \right)$ cái dòng cuối là $x^2$ rồi làm sao để ra $x^3$

$\sqrt[3]{x^2-1}+x=\sqrt{x^3-2}$

Cho $a,b,c$ là các số thực dương thoả mãn $a^2+b^2+c^2=1$. Chứng minh:

$\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}+\frac{b^2+bc+1}{\sqrt{b^2+3bc+c^2}}+\frac{c^2+ca+1}{\sqrt{c^2+3ca+b^2}}\geq \sqrt{5}(a+b+c)$