Lời giải:
Từ đk đề bài suy ra \(2x+2\geq 0\Rightarrow x\geq -1\)
\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
\(\Leftrightarrow \sqrt{2(x^2+4x+3)}+\sqrt{x^2-1}-2(x+1)=0\)
\(\Leftrightarrow \sqrt{2(x+1)(x+3)}+\sqrt{(x+1)(x-1)}-2(x+1)=0\)
\(\Leftrightarrow \sqrt{x+1}(\sqrt{2(x+3)}+\sqrt{x-1}-2\sqrt{x+1})=0\)
\(\Rightarrow \left[\begin{matrix} \sqrt{x+1}=0(1)\\ \sqrt{2(x+3)}+\sqrt{x-1}-2\sqrt{x+1}=0(2)\end{matrix}\right.\)
Với \((1)\Rightarrow x=-1\) (t.m)
Với \((2)\Leftrightarrow \frac{2(x+3)-4(x+1)}{\sqrt{2(x+3)}+2\sqrt{x+1}}+\sqrt{x-1}=0\)
\(\Leftrightarrow \frac{2(1-x)}{\sqrt{2(x+3)}+2\sqrt{x+1}}+\sqrt{x-1}=0\)
\(\Leftrightarrow \sqrt{x-1}(1-\frac{2\sqrt{x-1}}{\sqrt{2(x+3)}+2\sqrt{x+1}})=0\)
Dễ thấy \(0\leq 2\sqrt{x-1}< \sqrt{2(x+3)}+2\sqrt{x+1}\Rightarrow \frac{2\sqrt{x-1}}{\sqrt{2(x+3)}+2\sqrt{x+1}}< 1\)
Do đó \(\sqrt{x-1}=0\Rightarrow x=1\) (t.m)
Vậy \(x=\pm 1\)
