Đáp án + giải thích các bước giải:
$ĐKXD: \begin{cases} x^2-1\ge0(1) \\ x^2+x\ge0(2) \\(x+1)(2x+3)\ge0(3) \end{cases} $
`(1)->x^2-1>=0`
`->x^2>=1`
`->`\(\left[ \begin{array}{l}x\ge1\\x\le-1\end{array} \right.\)
`(2)->x^2+x>=0`
`->x(x+1)>=0`
`->`\(\left[ \begin{array}{l}x\ge0\\x\le-1\end{array} \right.\)
`(3)->(x+1)(2x+3)>=0`
`->`\(\left[ \begin{array}{l}\begin{cases} x+1\ge0 \\ 2x+3\ge0 \end{cases} \\\begin{cases} x+1\le0 \\2x+3\le0\end{cases}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}\begin{cases} x\ge-1 \\ x\ge\dfrac{-3}{2} \end{cases} \\\begin{cases} x\le-1 \\x\le\dfrac{-3}{2}\end{cases}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x\ge-1\\x\le-\dfrac{3}{2}\end{array} \right.\)
`\sqrt{x^2-1}+\sqrt{x^2+x}=\sqrt{(x+1)(2x+3)}`
`->\sqrt{(x-1)(x+1)}+\sqrt{x(x+1)}-\sqrt{(x+1)(2x+3)}=0`
`->\sqrt{x+1}(\sqrt{x-1}+\sqrt{x}-\sqrt{2x+3})=0`
Xét `\sqrt{x+1}=0`
`->x+1=0`
`->x=-1(TM)`
Xét `\sqrt{x-1}+\sqrt{x}-\sqrt{2x+3}=0`
`->\sqrt{x-1}+\sqrt{x}=\sqrt{2x+3}`
`->x-1+x+2\sqrt{x(x-1)}=2x+3`
`->2\sqrt{x^2-x}=4`
`->\sqrt{x^2-x}=2`
`->x^2-x=4`
`->x^2-x-4=0`
`Δ=17`
`->`\(\left[ \begin{array}{l}x=\dfrac{1+\sqrt{17}}{2}\\x=\dfrac{1-\sqrt{17}}{2}\end{array} \right.(TM)\)
Vậy `S={-1;(1\pm\sqrt{17})/2}`
