1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

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

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

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

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

2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)

\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)

\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)

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

\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)

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

\(=4\sqrt{5}\)

\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)

\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)

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

b)\(=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c)\(=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2\sqrt{5}-3-\sqrt{5}+2\)
\(=\sqrt{5}-1\)

b,\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)

c,\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(3-2\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}=3-2\sqrt{5}-\sqrt{5}+2=5-3\sqrt{5}\)

a: \(=\left(\sqrt{3}-2\right)\cdot\sqrt{\left(2+\sqrt{3}\right)^2}\)

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

=3-4=-1

b: \(=\sqrt{6+4\sqrt{2}}-\sqrt{11-2\sqrt{18}}\)

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

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

c: \(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)

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

\(=4\sqrt{5}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{7}+\sqrt{3}\right|-\left|\sqrt{7}-\sqrt{3}\right|\right)\)

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

\(\sqrt{5+\sqrt{21}}-\sqrt{5-\sqrt{21}}\\ =\dfrac{\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)}{\sqrt{2}}\\ =\dfrac{\left(\sqrt{7+2\sqrt{7}.\sqrt{3}+3}-\sqrt{7-2\sqrt{7}.\sqrt{3}+3}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{3}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

\(=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{3}{2}}=2\sqrt{\dfrac{3}{2}}=\sqrt{6}\)

 \(=\sqrt{5.\left(\sqrt{3}+1\right)}.\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)

\(=\sqrt{5}.\left(\sqrt{3}+1\right).\sqrt{48-10.\left(2+\sqrt{3}\right)}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{28-10\sqrt{3}}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{\left(5-\sqrt{3}\right)^2}\)

\(=\left(\sqrt{15}+\sqrt{5}\right).\left(5-\sqrt{3}\right)\)

Vậy...

~ Chắc chắn đúng cậu nhé ~ Tiếc gì 1 tk cho tớ nào?

\(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}=\dfrac{3-\sqrt{5}}{2}\)

a)\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\sqrt{\dfrac{1}{2}\left(16+8\sqrt{3}\right)}-\sqrt{\dfrac{1}{2}\left(16-8\sqrt{3}\right)}\)

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

b)\(=\dfrac{\sqrt{16+2.4\sqrt{5}+5}}{4+\sqrt{5}}.\sqrt{\left(2-\sqrt{5}\right)^2}\)\(=\dfrac{\sqrt{\left(4+\sqrt{5}\right)^2}}{4+\sqrt{5}}\left|2-\sqrt{5}\right|=\sqrt{5}-2\)

a) Ta có: \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)

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

\(=2\sqrt{2}\)

b) Ta có: \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}\cdot\sqrt{9-4\sqrt{5}}\)

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

=16-5=11