GPT: (ĐKXĐ cụ thể giúp em vs ạ) $\sqrt[]{2x^2+10x+8}$ + $\sqrt[]{x^2-1}$ = $2x+2$

`sqrt{2x^2+10x+8}+sqrt{x^2−1}=2x+2`
`ĐKXĐ`
$\begin{cases}2x^2+10x+8\\x^2-1\\\end{cases}$
`<=>`$\begin{cases}2x^2+2x+8x+8\\(x-1)(x+1)\\\end{cases}$
`<=>`$\begin{cases}(2x+8)(x+1)\\(x-1)(x+1)\\\end{cases}$
`<=>$\begin{cases}\(\left[\begin{array}{l}x\geq-1 \\x\leq-4 \end{array} \right.\\\\(\left[ \begin{array}{l}x\geq1\\x \leq-1\end{array} \right.\) \\\end{cases}$

`-1<=x<=1`
`<=>sqrt{(2x+8)(x+1)}+sqrt{(x-1)(x+1)}=2(x+1)`
`sqrt{x+1}(sqrt{2x+8}+sqrt{x-1}+sqrt{x+1}=0`
`=>x=-1(TM)`
vì `sqrt{x+1}(sqrt{2x+8}+sqrt{x-1}+sqrt{x+1}=0`
dấu = xảy ra khi
`{x=-1`
`{x=-4`
`{x=1`
vô lý