ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow x^3+x+2+\sqrt{x^3+x+2}=x^4-7x^2+12\)
\(\Leftrightarrow x^3+x+2+\sqrt{x^3+x+2}=\left(x^2-4\right)^2+x^2-4\)
Đặt \(\left\{{}\begin{matrix}x^2-4=a\\\sqrt{x^3+x+2}=b\end{matrix}\right.\)
\(\Rightarrow b^2+b=a^2+a\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a+b+1=0\end{matrix}\right.\)
TH1: \(\sqrt{x^3+x+2}=x^2-4\)
\(\sqrt{x^3+x+2}=x^2-4\) (\(x\ge2\))
Rất tiếc pt này ko giải được, nghiệm rất xấu (dù có nghiệm) :)
TH2: \(\sqrt{x^3+x+2}+x^2-3=0\Leftrightarrow\sqrt{x^3+x+2}=3-x^2\) \(\left(x^2\le3\right)\)
\(\Leftrightarrow x^4-x^3-6x^2-x+7=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-6x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x^3-6x-7=0\end{matrix}\right.\)
Xét \(x^3-6x-7=0\Leftrightarrow\left(x+1\right)\left(x^2-3\right)-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}=0\)
Do \(\left\{{}\begin{matrix}x\ge-1\\x^2\le3\end{matrix}\right.\) \(\Rightarrow\left(x+1\right)\left(x^2-3\right)\le0\Rightarrow VT< 0\Rightarrow\) ptvn
