A = \(\frac{1+x}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{3}\)=\(\frac{1+x}{3\sqrt{x}}\)

ĐKXĐ : x > 0

\(A=\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\)

\(=\sqrt{\frac{2\left(5+\sqrt{21}\right)}{2}}+\sqrt{\frac{2\left(5-\sqrt{21}\right)}{2}}\)

\(=\sqrt{\frac{10+2\sqrt{21}}{2}}+\sqrt{\frac{10-2\sqrt{21}}{2}}\)

\(=\sqrt{\frac{3+2\sqrt{21}+7}{2}}+\sqrt{\frac{3-2\sqrt{21+7}}{2}}\)

\(=\sqrt{\frac{\left(\sqrt{3}+\sqrt{7}\right)}{2}}+\sqrt{\frac{\left(\sqrt{3}+\sqrt{7}\right)}{2}}\)

\(=\frac{\left|\sqrt{3}+\sqrt{7}\right|}{2}+\frac{\left|\sqrt{3}-\sqrt{7}\right|}{2}\)

\(=\frac{\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{7}}{2}\)

\(=\sqrt{\frac{6}{2}}=\sqrt{3}\)

\(\sqrt{x}=x\)      \(\left(x\ge0\right)\)

\(\Leftrightarrow\sqrt{x}-x=0\)

\(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\1-\sqrt{x}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}\left(tm\right)}\)

Vậy 

\(\hept{\begin{cases}x=0\\x=1\end{cases}}\)

6:2.3=3.3=9

\(6:2.(1+2)\)

\(=3.3\)

\(=9\)

~ Hok tốt a~

\(A=\sqrt{2012^2+2012^2.2013^2+2013^2}\)

     \(=\sqrt{2012^2+\left(2012.2013\right)^2+2013^2}\)

       \(=2012+2012.2013+2013\)

Vậy A là một số tự nhiên

P/s: Mình nghĩ thế, không chắc!

\(A=\sqrt{2012^2+2012^2.2013^2+2013^2}\)

\(=\sqrt{\left(2013-1\right)^2+2012^2.2013^2+2013^2}\)

\(=\sqrt{2.2013^2-2.2013+1+2012^2.2013^2}\)

\(=\sqrt{2.2013.\left(2013-1\right)+1+2012^2.2013^2}\)

\(=\sqrt{2012^2.2013^2+2.2013.2012+1}=\sqrt{\left(2012.2013+1\right)^2}=2012.2013+1\)

\(A=4-\sqrt{x^2-4x+4}\)

Ta có :  \(\sqrt{x^2-4x+4}\ge0\)

\(\Rightarrow4-\sqrt{x^2-4x+4}\le4\)

hay  \(A\le4\)

Dấu "=" xảy ra khi : 

\(x^2-4x+4=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy  \(A_{Max}=4\Leftrightarrow x=2\)

\(A=4-\sqrt{x^2-4x+4}\)

\(=4-\sqrt{\left(x-2\right)}^2\)

\(4-\left|x-2\right|\)

Vì \(\left|x-2\right|\ge0\)

\(\Rightarrow-\left|x-2\right|< 0\)

\(\Leftrightarrow4-\left|x-2\right|< 4\)

\(A\le4\)

Dấu " = " xảy ra khi x - 2 = 0 <=> x = 2

Vậy .....