\(\Leftrightarrow\frac{6x^2}{\sqrt{2x^2+1}+1}=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{6x}{\sqrt{2x^2+1}+1}=8\sqrt{2x^2+1}+3x+1\left(1\right)\end{matrix}\right.\)
Xét (1):
- Với \(x>0\Rightarrow\frac{6x}{\sqrt{2x^2+1}+1}=\frac{6}{\sqrt{2+\frac{1}{x^2}}+\frac{1}{x}}< \frac{6}{\sqrt{2}}< 5\)
\(\sqrt{2x^2+1}>1\Rightarrow8\sqrt{2x^2+1}+3x+1>8>5\)
\(\Rightarrow VT< VP\) pt vô nghiệm
- Với \(x< 0\Rightarrow VT=\frac{6x}{\sqrt{2x^2+1}+1}< 0\)
\(\sqrt{2x^2+1}>\sqrt{x^2}=\left|x\right|\Rightarrow VP>8\left|x\right|+3x+1>3\left|x\right|+3x+1>1>0\)
\(\Rightarrow VT< VP\) pt vô nghiệm
Vậy pt có nghiệm duy nhất \(x=0\)
