$Đk:x\ge0;x\ne1$

$a)\ A=\bigg(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\bigg).\dfrac{(1-x)^2}{2}\\=\dfrac{(2-\sqrt{x})(1+\sqrt{x})-(\sqrt{x}+2)(1-\sqrt{x})}{(1-\sqrt{x})(1+\sqrt{x})^2}.\dfrac{(1-x)^2}{2}\\=\dfrac{2+\sqrt{x}-x+\sqrt{x}+x-2}{(1-x)(1+\sqrt{x})}.\dfrac{(1-x)^2}{2}\\=\dfrac{2\sqrt{x}.(1-x)^2}{2(1-x)(1+\sqrt{x})}\\=\dfrac{\sqrt{x}(1-x)}{1+\sqrt{x}}\\=\sqrt{x}(1-\sqrt{x})$

$b)$ Để $A>0$ thì $\sqrt{x}(1-\sqrt{x})>0$

$\left\{\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x>0\\0\le x<1\end{matrix}\right.$

$\Leftrightarrow 0<x<1$

Vậy với $0<x<1$ thì $A>0$

$c)\ A=\sqrt{x}(1-\sqrt{x})\\=\sqrt{x}-x\\=\dfrac{1}{4}-\bigg(x-\sqrt{x}+\dfrac{1}{4}\bigg)\\=\dfrac{1}{4}-\bigg(\sqrt{x}-\dfrac{1}{2}\bigg)^2\le\dfrac{1}{4}$

Dấu $=$ xảy ra khi

$\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}$

Vậy $A_{max}=\dfrac{1}{4}$ khi $x=\dfrac{1}{4}$

Đáp án:

`A=((sqrtx-2)/(x-1)-(sqrtx+2)/(x+2sqrtx+1))(1-x)^2/2`

`a)đk:x>=0,x ne 1`

`A=((sqrtx-2)/((sqrtx-1)(sqrtx+1))-(sqrtx+2)/(sqrtx+1)^2)(x-1)^2/2`

`A=((sqrtx-2)(sqrtx+1)-(sqrtx+2)(sqrtx-1))/((sqrtx+1)^2(sqrtx-1))(x-1)^2/2`

`A=(x-sqrtx-2-x-sqrtx+2)/((x-1)(sqrtx+1))(x-1)^2/2`

`A=(-2sqrtx)/((x-1)(sqrtx+1))(x-1)^2/2`

`A=-sqrtx(sqrtx-1)`

`b)` để `A>0`

`<=>-sqrtx(sqrtx-1)>0`

`<=>sqrtx(sqrtx-1)<0`

Mà `sqrtx>=0`

`<=>` \(\begin{cases}\sqrt{x}>0\\\sqrt{x}-1<0\\\end{cases}\)

`<=>` \(\begin{cases}x>0\\x<1\\\end{cases}\)

`<=>0<x<1.`

`c)A=-sqrtx(sqrtx-1)`

`=-x+sqrtx`

`=-(x-sqrtx)`

`=-(x-sqrtx+1/4)+1/4`

`=-(sqrtx-1/2)^2+1/4<=1/4`

Dấu '=' xảy ra khi `sqrtx=1/2<=>x=1/4.`