ĐK: `{(2x^2+8x+6>=0),(x^2-1>=0),(2x+2>=0):} <=> {(x=-1),(x>=1):}`
`\sqrt(2x^2+8x+6)+\sqrt(x^2-1)=2x+2`
`<=>(2x^2+8x+6)+(x^2-1)+2\sqrt((2x^2+8x+6)(x^2-1))=(2x+2)^2`
`<=>2(x+3)(x+1)+(x-1)(x+2)+2\sqrt((x+1)^2 (x+3)(x-1))=4(x+1)^2`
`<=> (x+1)[2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0`
`<=> [(x=-1\ (TM)),([2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0\ (1)):}`
(1) `<=> x-1=2\sqrt((x+3)(x-1))`
`<=>x^2-2x+1=4(x+3)(x-1)`
`<=>x=1\ `(TM)
Vậy `S={\pm 1}`.
\(ĐK:x\le-3;x\ge-1\)
\(PT\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\left(x+3\right)+\left(x-1\right)+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\\ \Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\\ \Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{25}{7}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)
Vậy \(S=\left\{-1;1\right\}\)
