a)

\(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{5+2\sqrt{15}+3}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}\\ =\left(\sqrt{9}-\sqrt{15}\right)\cdot\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}\)

\(=\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\) (vì \(\sqrt{5}+\sqrt{3}>0\))

\(=\sqrt{3}\cdot\dfrac{3-5}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-2}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-\sqrt{4}}{\sqrt{2}}\\ =-\sqrt{6}\)

b)

\(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\\ =\sqrt{20-2\cdot3\cdot2\sqrt{5}+9}-\sqrt{20-2\cdot2\cdot2\sqrt{5}+4}\\ =\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}-3\right|-\left|2\sqrt{5}-2\right|\)

\(=2\sqrt{5}-3-\left(2\sqrt{5}-2\right)\) (vì \(2\sqrt{5}-3>0;2\sqrt{5}-2>0\))

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

a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)

\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)

\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)

\(=14\sqrt{3}-2\sqrt{57}\)

b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)

\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)

\(=\left(4-6+3-5\right)\sqrt{6}\)

\(=-4\sqrt{6}\)

a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

\(=2\sqrt{3}\)

c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)

a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)

  \(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)

  \(=0\)

b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)

Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)

\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)

Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)

a) E=0

b) F=căn 3/2

9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)

10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)

12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)

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

=9-3=6

13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)

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

a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)

b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}\)

\(=\dfrac{\sqrt{6}}{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\)

\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=\dfrac{6-\sqrt{6}}{2}\)

c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)

d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)

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

e, Ghi đúng đề.

\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)

\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)

\(T=\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{\frac{35}{4}}=\left\{\left[\left(\frac{b}{a}\right)^{-1}\left(\frac{b}{a}\right)^{\frac{1}{5}}\right]^{\frac{1}{7}}\right\}^{\frac{35}{4}}=\left[\left(\frac{b}{a}\right)^{-\frac{4}{5}}\right]=\frac{a}{b}\)

\(T=\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{\frac{35}{4}}=\sqrt[4]{\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{35}}=\sqrt[4]{\left(\frac{a}{b}\sqrt[5]{\frac{b}{a}}\right)^5}\)

\(=\sqrt[4]{\left(\frac{a}{b}\right)^5.\frac{b}{a}}=\sqrt[4]{\left(\frac{a}{b}\right)^4}=\frac{a}{b}\)