$\sqrt[]{x^2 +x}$ + $\sqrt[]{x-2x}$ =2$\sqrt[]{x^2}$

`sqrt{x^2+x}+sqrt{x-2x}=2sqrt{x^2}`

`<=>sqrt{x(x+1)}+sqrt{-x}=2|x|`

`đkxđ`:$\begin{cases}x(x+1) \ge 0\\-x \ge 0\\\end{cases}$

`<=>` $\begin{cases}x \le 0\\x+1 \le 0\\\end{cases}$

`<=>` $\begin{cases}x \le 0\\x \le -1\\\end{cases}$

`<=>x<=-1`

`pt<=>sqrt{x(x+1)}+sqrt{-x}=-2x`

`<=>sqrt{x(x+1)}+sqrt{-x}+2x=0`

`<=>sqrt{-x}(sqrt{-(x+1)}+1+sqrt{-x})=0`

Vì `sqrt{-(x+1)}+1+sqrt{-x}>0AAx<=-1`

`<=>sqrt{-x}=0`

`<=>x=0`(loại vì `x<=-1`)

Vậy phương trình vô nghiệm.