Tìm x,biết $\sqrt[]{x+4}$ - $\sqrt[]{1-x}$ = $\sqrt[]{1-2x}$

`#huy`

`\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}`

`<=>\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}`

`<=>{(-4<=x<=1/2),(x+4=1-x+2\sqrt{(1-x)(1-2x)}+1-2x):}`

`<=>{(-4<=x<=1/2),(\sqrt{(1-x)(1-2x)}=2x+1):}`

`<=>{(-4<=x<=1/2),(x>=1/2),((1-x)(1-2x)=4x^2+4x+1):}`

`<=>{(-1/2<=x<=1/2),(x=0vvx=-7/2):}`

`<=>x=0`

Đáp án: $x=0$

Giải thích các bước giải:

$\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}$   ĐK: $-4\le x\le \dfrac{1}{2}$
$\Leftrightarrow \left( \sqrt{x+4}-2 \right)+\left( 1-\sqrt{1-x} \right)+\left( 1-\sqrt{1-2x} \right)=0$
$\Leftrightarrow \dfrac{x}{\sqrt{x+4}+2}+\dfrac{x}{1+\sqrt{1-x}}+\dfrac{2x}{1+\sqrt{1-2x}}=0$
$\Leftrightarrow x\left( \dfrac{1}{\sqrt{x+4}+2}+\dfrac{1}{1+\sqrt{1-x}}+\dfrac{2}{1+\sqrt{1-2x}} \right)=0$
$\Leftrightarrow x=0$ (thỏa mãn) (vế sau luôn dương)