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Answer 1

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

\(B=\frac{\sqrt{4x^2-12x+9}}{2x-3}=\frac{\sqrt{(2x-3)^2}}{2x-3}=\frac{|2x-3|}{2x-3}=\frac{-(2x-3)}{2x-3}=-1\) (do $2x-3<0$)

\(C=20\sqrt{2}+4\sqrt{2}-24\sqrt{2}-4\sqrt{2}=\sqrt{2}(20+4-24-4)=-4\sqrt{2}\)

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

\(=\frac{4}{\sqrt{2^2-3}}=4\)

Answer 2

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

$=|\sqrt{x-4}+2|=\sqrt{x-4}+2$
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$H\sqrt{2}=\sqrt{2x+4\sqrt{2x-4}}=\sqrt{(2x-4)+4\sqrt{2x-4}+4}$

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

$\Rightarrow H=\sqrt{x-2}+\sqrt{2}$

Answer 3

\(E=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{-(\sqrt{2}-1)}+\frac{\sqrt{5}(\sqrt{3}-1)}{-(\sqrt{3}-1)}\right].(\sqrt{7}-\sqrt{5})\)

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

\(F\sqrt{2}=\sqrt{12-2\sqrt{11}}-\sqrt{12+2\sqrt{11}}=\sqrt{11-2\sqrt{11}+1}-\sqrt{11+2\sqrt{11}+1}\)

\(=\sqrt{(\sqrt{11}-1)^2}-\sqrt{(\sqrt{11}+1)^2}=|\sqrt{11}-1|-|\sqrt{11}+1|=-2\)

$\Rightarrow F=-\sqrt{2}$