Đặt \(\sqrt{x^2+3}=a\ge\sqrt{3}\) (1)
pt \(\Leftrightarrow\left(a^2-3\right)^2+a-3=0\)
\(\Leftrightarrow a^4+9-6a^2+a-3=0\)
\(\Leftrightarrow a^4-4a^2-2a^2+4a-3a+6=0\)
\(\Leftrightarrow\left(a-2\right)\left(a^3+2a^2-2a-3=0\right)\)
\(\Leftrightarrow\left(a-2\right)\left(a^3+a^2+a^2+a-3a-3\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+1\right)\left(a^2+a-3\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+1\right)\left[\left(a+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}a-2=0\\a+1=0\\\left(a+\dfrac{1}{2}\right)^2-\dfrac{13}{4}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(c\right)\\a=-1\left(l\right)\\a=\dfrac{-1+\sqrt{13}}{2}\left(l\right)\\a=\dfrac{-1-\sqrt{13}}{2}\left(l\right)\end{matrix}\right.\)
Thay a = 2 vào (1) ta được: \(\sqrt{x^2+3}=2\Rightarrow x^2+3=4\)
\(\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
Vây phương trình có nghiêm là x=1 hay x=-1