a) \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\) = \(\dfrac{\sqrt{10+4\sqrt{6}}}{\sqrt{2}}+\dfrac{\sqrt{10-4\sqrt{6}}}{\sqrt{2}}\)

= \(\dfrac{\sqrt{\left(\sqrt{6}+2\right)^2}}{\sqrt{2}}+\dfrac{\sqrt{\left(\sqrt{6}-2\right)^2}}{\sqrt{2}}\) = \(\dfrac{\sqrt{6}+2}{\sqrt{2}}+\dfrac{\sqrt{6}-2}{\sqrt{2}}\)

= \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}}+\dfrac{\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{2}}\) = \(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\)

= \(2\sqrt{3}\)

b) \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\) = \(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

= \(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\) = \(\dfrac{\sqrt{3}-1}{\sqrt{2}}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

= \(\dfrac{-2}{\sqrt{2}}\) = \(-\sqrt{2}\)

ừ

\(\left(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\right)+\left(\dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\right)\)

= \(\left(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\right)+\left(\dfrac{3-2\sqrt{2}-3-2\sqrt{2}}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\right)\)

= \(9\sqrt{2}+\left(\dfrac{-4\sqrt{2}}{1}\right)\) = \(9\sqrt{2}-4\sqrt{2}\) = \(5\sqrt{2}\)

a) đkxđ : x \(\ge\) 0

A = \(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\) = \(\dfrac{x^2+\sqrt{x}+x-\sqrt{x}+1}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)

= \(\dfrac{x^2+x+1}{x-\sqrt{x}+1}-\dfrac{2\sqrt{x}+1}{1}\) = \(\dfrac{x^2+x+1-\left(2\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)

= \(\dfrac{x^2+x+1-\left(2x\sqrt{x}-2x+2\sqrt{x}+x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)

= \(\dfrac{x^2+x+1-2x\sqrt{x}+2x-2\sqrt{x}-x+\sqrt{x}-1}{x-\sqrt{x}+1}\)

= \(\dfrac{x^2-2x\sqrt{x}+2x-\sqrt{x}}{x-\sqrt{x}+1}\) = \(\dfrac{\left(x-\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{x-\sqrt{x}+1}\) = \(x-\sqrt{x}\)

b) ta có A = 2 \(\Leftrightarrow\) \(x-\sqrt{x}=2\) \(\Leftrightarrow\) \(x-\sqrt{x}-2=0\)

giải phương trình ta có : x = 4

c) ta có đkxđ là \(x\ge0\) \(\Leftrightarrow\) \(x-\sqrt{x}\ge0\)

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)