Bài 1 :

\(A=\sqrt{4-2\sqrt{3}}+\sqrt{27}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{27}\)

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

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

\(=\sqrt{3}-1+3\sqrt{3}\)

\(=4\sqrt{3}-1\)

\(B=\sqrt{14-6\sqrt{5}}+\sqrt{125}\)

\(=\sqrt{9-6\sqrt{5}+5}+\sqrt{125}\)

\(=\sqrt{\left(3-\sqrt{5}\right)}^2+5\sqrt{5}\)

\(=\left|3-\sqrt{5}\right|+5\sqrt{5}\)

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

\(=3+4\sqrt{5}\)

\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\\ =\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\\ =\left(3\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\\ =21-2\sqrt{21}+2\sqrt{21}\\ =21\)

6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)

7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)

8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)

9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)

10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)

a: ĐKXĐ: x-10>=0

=>x>=10

b: \(\sqrt{9a^2b}=\sqrt{\left(3a\right)^2\cdot b}=3a\cdot\sqrt{b}\)

c: \(\left(2\sqrt{3}+1\right)^2=13+4\sqrt{3}\)

\(\left(2\sqrt{2}+\sqrt{5}\right)^2=8+5+2\cdot2\sqrt{2}\cdot\sqrt{5}=13+4\sqrt{10}\)

mà \(4\sqrt{3}< 4\sqrt{10}\left(3< 10\right)\)

nên \(\left(2\sqrt{3}+1\right)^2< \left(2\sqrt{2}+\sqrt{5}\right)^2\)

=>\(2\sqrt{3}+1< 2\sqrt{2}+\sqrt{5}\)