Đáp án:
Giải thích các bước giải:
`\sqrt{x(x-1)} + \sqrt{x(x+2)} = 2\sqrt{x^{2}}`
`ĐKX Đ : {(x(x-1) >=0),(x(x+2) >=0):} <=>`$\begin{cases} \left[ \begin{array}{l}x \le 0\\x \ge 1\end{array} \right. \\\left[ \begin{array}{l}x\le-2\\x\ge0\end{array} \right. \end{cases}$
`<=>`\(\left[ \begin{array}{l}x = 0\\x \le -2\\x \ge 0\end{array} \right.\) `(1)`
`(**) <=>x(x-1)+x(x+2)+2\sqrt{x^{2}(x-1)(x+2)} = 4x^{2}`
`<=>2\sqrt{x^{2}(x^{2}+x-2)} = 2x^{2} - x = x(2x-1)`
`<=>`$\begin{cases} \left[ \begin{array}{l}x\le0\\x\ge\dfrac{1}{2}\end{array} \right.\\4x^{2}(x^{2}+x-2)=x^{2}(2x-1)^{2} \end{cases}$ `<=>`$\begin{cases} \left[ \begin{array}{l}x\le0\\x\ge\dfrac{1}{2}\end{array} \right.\\x^{2}[4(x^{2}+x-2)-(2x-1)^{2}]=0 \end{cases}$
`<=>`$\begin{cases} \left[ \begin{array}{l}x\le0\\x\ge\dfrac{1}{2}\end{array} \right.\\x^{2}(8x-9) = 0 \end{cases}$ `<=>`$\begin{cases} \left[ \begin{array}{l}x\le0\\x\ge\dfrac{1}{2}\end{array} \right.\\ \left[ \begin{array}{l}x^{2} = 0\\x=\dfrac{9}{8}\end{array} \right.\end{cases}$
Vậy `x = {0;9/8}`
