ĐKXĐ: \(x\ge-3\)
\(x^2+1+2\left(x+3\right)-3\sqrt{\left(x^2+1\right)\left(x+3\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+1}=a>0\\\sqrt{x+3}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2+2b^2-3ab=0\)
\(\Rightarrow\left(a-b\right)\left(a-2b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=\sqrt{x+3}\\\sqrt{x^2+1}=2\sqrt{x+3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+1=x+3\\x^2+1=2\left(x+3\right)\end{matrix}\right.\)
\(\Rightarrow...\) (đối chiếu nghiệm với ĐKXĐ)
\(< =>\left(x^2+2x+7\right)^2=9\left(x^2+1\right)\left(x+3\right)\)\(\left(x\ge-3\right)\)
\(< =>x^4+4x^3+18x^2+28x+49=9x^3+27x^2+9x+27\)
\(< =>x^4-5x^3-9x^2+19x+22=0\)
\(< =>\left(x+2\right)\left(x-1\right)\left(x^2-4x-11\right)=0\)
\(=>\left[{}\begin{matrix}x=-2\\x=1\\\left\{{}\begin{matrix}x=2-\sqrt{15}\\x=2+\sqrt{15}\end{matrix}\right.\end{matrix}\right.\)(tm)
