Đáp án + giải thích các bước giải:
$ĐKXD: \begin{cases} 1-x^2\ge0(1) \\ x^2+3x+2\ge0(2) \\ x+1\ge0 (3) \end{cases} $
`(1)->x^2<=1`
`->-1<=x<=1`
`(2)->x^2+x+2x+2>=0`
`->(x+1)(x+2)>=0`
`->`\(\left[ \begin{array}{l}x\ge-1\\x\le-2\end{array} \right.\)
`(3) ->x>=-1`
`\sqrt{1-x^2}+\sqrt{x^2+3x+1}=x+1`
`->\sqrt{(1-x)(1+x)}+\sqrt{x^2+x+2x+2}=x+1`
`->\sqrt{(1-x)(1+x)+\sqrt{x(x+1)+2(x+1)}=x+1`
`->\sqrt{(1-x)(1+x)+\sqrt{(x+1)(x+2)}-(x+1)=0`
`->\sqrt{x+1}(\sqrt{1-x}+\sqrt{x+2}-\sqrt{x+1})=0`
Xét `\sqrt{x+1}=0`
`->x+1=0`
`->x=-1(TM)`
Xét `\sqrt{1-x}+\sqrt{x+2}-\sqrt{x+1}=0`
`->\sqrt{1-x}+\sqrt{x+2}=\sqrt{x+1}`
`->1-x+x+2+2\sqrt{(1-x)(x+2)}=x+1`
`->2\sqrt{x-x^2+2-2x}=x-2 (x>=2)`
`->4(-x^2-x+2)=x^2-4x+4`
`->-4x^2-4x+8=x^2-4x+4`
`->5x^2=4`
`->x^2=4/5`
mà `x>=2->x^2>=4 `
`->`Vô nghiệm
Vậy `S={-1}`
