ĐKXĐ: \(x\ge-2\)
Pt cho \(\Leftrightarrow4\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}=x^2+x+10\)
Đặt \(\sqrt{x+2}=a;\sqrt{x^2-2x+4}=b\left(a,b\ge0\right)\)
Khi đó ta được pt: \(4ab=b^2+3a^2\Leftrightarrow\left(b-a\right)\left(b-3a\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\b=3a\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x+2}=\sqrt{x^2-2x+4}\left(1\right)\\\sqrt{x^2-2x+4}=3\sqrt{x+2}\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tm\right)}\)
\(\left(2\right)\Leftrightarrow x^2-11x-14=0\Leftrightarrow\left(x-\frac{11}{2}\right)^2=\frac{177}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11+\sqrt{177}}{2}\\x=\frac{11-\sqrt{177}}{2}\end{cases}\left(tm\right)}\)
Vậy tập nghiệm của pt là \(S=\left\{1;2;\frac{11\pm\sqrt{177}}{2}\right\}.\)
