\(\sqrt{7-2\sqrt{12}}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

=> Chọn C

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, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)

       = \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)

       = \(\frac{-2\sqrt{6}}{2}\)

       = \(-\sqrt{6}\)

a, =\(9\sqrt{2}\)

b, =21

a) \(=9\sqrt{2}\)

b) \(=21\)

học tốt.

c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

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

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)

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) A=12\(\sqrt{3}\)

    B= \(\frac{8}{3}\)

c) C= 1

d)...

Chúc bạn học tốt nha ^^!