\(M=\dfrac{3}{2}\cdot4\sqrt{2x}-\dfrac{1}{3}\cdot3\sqrt{2x}+\dfrac{2}{5}\cdot5\sqrt{2x}-4\sqrt{2x}=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)

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

2: \(=\sqrt{5}+2-\sqrt{5}=2\)

Để A>-2 thì \(-x+\sqrt{x}+2>0\)

\(\Leftrightarrow x-\sqrt{x}-2>0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)>0\)

=>\(\sqrt{x}-2>0\)

=>x>4

\(=4-\dfrac{12}{5}\sqrt{5}-6+\dfrac{18}{5}\sqrt{5}+2\sqrt{5}-6\)

\(=-8+\dfrac{16}{5}\sqrt{5}\)

\(2\sqrt{48}-3\sqrt{75}+5\sqrt{3}\)

\(=2\sqrt{16.3}-3\sqrt{25.3}+5\sqrt{3}\)

\(=2\sqrt{4^2.3}-3\sqrt{5^2.3}+5\sqrt{3}\)

\(=2.4\sqrt{3}-3.5\sqrt{3}+5\sqrt{3}\)

\(=8\sqrt{3}-15\sqrt{3}+5\sqrt{3}\)

\(=\left(8-15+5\right).\sqrt{3}\)

\(=-2\sqrt{3}\)

\(=\sqrt{8}\cdot\sqrt{8}-\sqrt[3]{\dfrac{16}{2}}=8-\sqrt[3]{8}=8-2=6\)

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

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