a) \(3\sqrt{2x}-4\sqrt{2x}+8-2\sqrt{x}\)

\(=-\left(4\sqrt{2x}-3\sqrt{2x}\right)+8-2\sqrt{x}\)

\(=-\sqrt{2x}-2\sqrt{x}+8\) 

b) \(3\sqrt{2x}-\sqrt{72x}+3\sqrt{18x}+18\)

\(=3\sqrt{2x}-6\sqrt{2x}+3\cdot3\sqrt{2x}+18\)

\(=3\sqrt{2x}-6\sqrt{2x}+9\sqrt{2x}+18\)

\(=\left(3+9-6\right)\sqrt{2x}+18\)

\(=6\sqrt{2x}+18\)

ĐKXĐ: \(x\ge2\)

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

\(=\sqrt{x-2+2.\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2.\sqrt{x-2}.\sqrt{2}+2}\)

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

\(=\left|\sqrt{x-2}+\sqrt{2}\right|+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)

Xét \(x\ge4\Rightarrow\sqrt{x-2}\ge\sqrt{2}\)

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

Xét \(0\le x< 4\Rightarrow\sqrt{x-2}< \sqrt{2}\)

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

\(S=\frac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x-\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-3x\sqrt{x}+2x-\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(S=\frac{x\sqrt{x}-2x+2\sqrt{x}-1+2x\sqrt{x}+x-2\sqrt{x}-1-3x\sqrt{x}+2x-\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(S=\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(S=\frac{1}{\sqrt{x}+1}\)

Vậy    \(S=\frac{1}{\sqrt{x}+1}\)

\(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3.\sqrt{5}}-\sqrt{2}\)

\(\sqrt{2}.A=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{9-2.3.\sqrt{5}+5}-2\)

\(\sqrt{2}.A=\sqrt{5}+1+3-\sqrt{5}-2=2\)

\(\Rightarrow A=\sqrt{2}\)

ĐKXĐ: \(\hept{\begin{cases}2x-4\ge0\\x+2.\sqrt{2x-4}\ge0\\x-2\sqrt{2x-4}\end{cases}}\Leftrightarrow x\ge2\)

\(\sqrt{x+2.\sqrt{2x-4}}+\sqrt{x-2.\sqrt{2x-4}}\)

\(=\sqrt{x-2+2.\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2.\sqrt{x-2}.\sqrt{2}+2}\)

\(=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)

Tự phá trị tuyệt đối

ĐKXĐ: 

\(2x-4\ge0\)và \(x+2\sqrt{2x-4}\ge0\)và \(x-2\sqrt{2x-4}\ge0\)

<=>\(2x\ge4\)và \(x\ge2\sqrt{2x-4}\)

<=>\(x\ge2\text{ và }x^2\ge8x-16\)

<=>\(x\ge2\text{ và }\left(x-4\right)^2\ge0\)

=>\(x\ge2\)

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

\(=\sqrt{x-2+2.\sqrt{2}\sqrt{x-2}+2}+\sqrt{x-2-2.\sqrt{2}\sqrt{x-2}+2}\)

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

\(=\left|\sqrt{x-2}+2\right|+\left|\sqrt{x-2}-2\right|\)

Với \(\sqrt{x-2}-2>0\) thì \(A=\sqrt{x-2}+2+\sqrt{x-2}-2=2\sqrt{x-2}\)

Với \(\sqrt{x-2}-2

\(\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}\)

\(=\sqrt{\frac{x-2\sqrt{2x-4}}{2}}\)

\(=\sqrt{\frac{x}{2}-\frac{2\sqrt{2x-4}}{2}}\)

\(=\sqrt{\frac{x}{2}-\sqrt{2x-4}}\)

\(=\sqrt{\frac{x}{2}-\sqrt{2x-4}}\)

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)