Đáp án:
`S={0;9/8}`
Giải thích các bước giải:
`\qquad \sqrt{x(x+2)}+\sqrt{x(x-1)}=2\sqrt{x^2}` $(1)$
$ĐK: \begin{cases}x(x+2)\ge 0\\x(x-1)\ge 0\\x^2\ge 0\ (luôn đúng)\end{cases}$
`<=>`$\left\{\begin{matrix}\left[\begin{array}{l}x\ge 0\\x\le -2\end{array}\right.\\\left[\begin{array}{l}x\ge 1\\x\le 0\end{array}\right.\end{matrix}\right.$
`<=>`$\left[\begin{array}{l}x\le -2\\x\ge 1\\x=0\end{array}\right.$
$\\$
`(1)<=>(\sqrt{x(x+2)}+\sqrt{x(x-1)})^2=(2\sqrt{x^2})^2`
`<=>x(x+2)+x(x-1)+2\sqrt{x^2 (x+2)(x-1)}=4x^2`
`<=>x^2+2x+x^2-x+2\sqrt{x^2(x^2+x-2)}=4x^2`
`<=>2\sqrt{x^2.(x^2+x-2)}=2x^2-x`
`<=>`$\begin{cases}2x^2-x\ge 0\\4x^2(x^2+x-2)=(2x^2-x)^2\end{cases}$
`<=>`$\left\{\begin{matrix}\left[\begin{array}{l}x\ge 0\\x\le \dfrac{1}{2}\end{array}\right.\\4x^2(x^2+x-2)=x^2 (2x-1)^2\ (2)\end{matrix}\right.$
`(2)<=>4x^2(x^2+x-2)-x^2.(4x^2-4x+1)=0`
`<=>x^2 .(4x^2+4x-8-4x^2+4x-1)=0`
`<=>x^2 .(8x-9)=0`
`<=>`$\left[\begin{array}{l}x^2=0\\8x-9=0\end{array}\right.$
`<=>`$\left[\begin{array}{l}x=0\ (thỏa\ đk)\\x=\dfrac{9}{8}\ (thỏa\ đk)\end{array}\right.$
Vậy phương trình có tập nghiệm: `S={0;9/8}`
