Lời giải:
PT \(\Leftrightarrow 3x^4-4x^3+(\sqrt{(x^2+1)^3}-1)=0\)
\(\Leftrightarrow x^2(3x^2-4x)+(\sqrt{x^2+1}-1)(x^2+1+\sqrt{x^2+1}+1)=0\)
\(\Leftrightarrow x^2(3x^2-4x)+\frac{x^2}{\sqrt{x^2+1}+1}(x^2+2+\sqrt{x^2+1})=0\)
\(\Leftrightarrow x^2\left[3x^2-4x+\frac{x^2+2+\sqrt{x^2+1}}{\sqrt{x^2+1}+1}\right]=0\)
Xét 2 TH:
TH1: $x^2=0\Leftrightarrow x=0$ (thỏa mãn)
TH2: \(3x^2-4x+\frac{x^2+2+\sqrt{x^2+1}}{\sqrt{x^2+1}+1}=0\)
Đặt $\sqrt{x^2+1}=a(a\geq 1)$ thì:
\(\frac{x^2+2+\sqrt{x^2+1}}{\sqrt{x^2+1}+1}=\frac{a^2+1+a}{a+1}=a+\frac{1}{a+1}\\ =\frac{a+1}{4}+\frac{1}{a+1}+\frac{3(a+1)}{4}-1\\ \geq 2\sqrt{\frac{1}{4}}+\frac{3(1+1)}{4}-1=\frac{3}{2}\) (áp dụng BĐT Cô-si)
\(\Rightarrow 3x^2-4x+\frac{x^2+2+\sqrt{x^2+1}}{\sqrt{x^2+1}+1}\geq 3x^2-4x+\frac{3}{2}=3(x-\frac{2}{3})^2+\frac{1}{6}>0\)
Suy ra TH 2 này không thỏa mãn
Vậy $x=0$
