\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

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

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

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

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)^2}\)

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

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

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

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

\(A=-3\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)}\)

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

\(B=1\)

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

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

\(=-3\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)}\)

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

\(=1\)

a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)

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

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

b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)

= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)

= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)

= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)

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.

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

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

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

yêu cầu ?

= 2 căn 6 - 3 + 3 - căn 6

= căn 6

Cho mình xin 1 tick nha trang

\(\sqrt{15-\sqrt{216}+\sqrt{33-12\sqrt{6}}}\)

\(=\sqrt{15-6\sqrt{6}+\sqrt{\left(2\sqrt{3}\right)^2-2\cdot2\sqrt{2}\cdot3+3^2}}\)

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

\(=\sqrt{15-6\sqrt{6}+\left|2\sqrt{3}-3\right|}\)

\(=\sqrt{15-6\sqrt{6}+2\sqrt{3}-3}\)

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

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

1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)

2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)

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

1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)

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

2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)

\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)

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

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