Đáp án:
Điều kiện:\(\begin{cases}x^2-x \ge 0\\x^2+x-2 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}x(x-1) \ge 0\\(x-1)(x+2) \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}\left[ \begin{array}{l}x \ge 1\\x \le 0\end{array} \right.\\\left[ \begin{array}{l}x \ge 1\\x \le -2\end{array} \right.\\\end{cases}\)
`<=>` \(\left[ \begin{array}{l}x \ge 1\\x \le -2\end{array} \right.\)
Với `x>=1`
`<=>\sqrt{x(x-1)}=\sqrtx.\sqrt{x-1},\sqrt{(x-1)(x+2)}=\sqrt{x-1}.\sqrt{x+2}`
`\sqrt{x^2-x}+\sqrt{x^2+x-2}=0`
`<=>\sqrtx.\sqrt{x-1}+\sqrt{x-1}.\sqrt{x+2}=0`
`<=>\sqrt{x-1}(\sqrt{x}+\sqrt{x+2})=0`
Vì `\sqrtx+\sqrt{x+2}>0AAx>=1`
`<=>\sqrt{x-1}=0`
`<=>x=1`.
Với `x<=-2`
`<=>\sqrt{x(x-1)}=\sqrt{-x}.\sqrt{1-x},\sqrt{(x-1)(x+2)}=\sqrt{1-x}.\sqrt{-x-2}`
`\sqrt{x^2-x}+\sqrt{x^2+x-2}=0`
`<=>\sqrt{-x}.\sqrt{1-x}+\sqrt{1-x}.\sqrt{-x-2}=0`
`<=>\sqrt{1-x}(\sqrt{-x}+\sqrt{-x-2})=0`
Vì `\sqrt{-x}+\sqrt{-x-2}>0AAx<=-2`
`<=>\sqrt{1-x}=0`
`<=>x=1`.
Vậy phương trình có nghiệm duy nhất `x=1`.
