KiritoNguyen

 

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 $x^3$