a ) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)

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

c ) \(\sqrt{21+4\sqrt{5}}=\sqrt{\left(2\sqrt{5}+1\right)^2}=2\sqrt{5}+1\)

d ) \(\sqrt{11+4\sqrt{7}}=\sqrt{\left(\sqrt{7}+2\right)^2}=\sqrt{7}+2\)

a) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)

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

c) \(\sqrt{21+4\sqrt{5}}=\sqrt{\left(2\sqrt{5}+1\right)^2}=2\sqrt{5}+1\)

d) \(\sqrt{11+4\sqrt{7}}=\sqrt{\left(\sqrt{7}+2\right)^2}=\sqrt{7}+2\)

a

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

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

b,

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

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

c,\(\sqrt{11+4\sqrt{7}}=\sqrt{11+2\sqrt{28}}=\sqrt{\sqrt{7}^2+2.\sqrt{7}.\sqrt{4}+\sqrt{4}^2}\)

\(=\sqrt{\left(\sqrt{7}+\sqrt{4}\right)^2}=\sqrt{7}+\sqrt{4}\)

a) \(\sqrt{5-2\sqrt{6}}=25\sqrt{-2\sqrt{6}}\)

b) \(\sqrt{3-2\sqrt{2}}=9\sqrt{-2\sqrt{2}}\)

c) \(\sqrt{11+4\sqrt{7}}=121\sqrt{4\sqrt{7}}\)

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

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

\(=4\sqrt{5}\)

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

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

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

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

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

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

\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\sqrt{20^2}-2.\sqrt{20}+1}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\left(\sqrt{20}-1\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-\left|\sqrt{20}-1\right|}}\)

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

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

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

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

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

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

\(=1\)

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

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

1, \(\sqrt{8+2\sqrt{15}}=\sqrt{8+2\sqrt{5.3}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)

2, \(\sqrt{15-2\sqrt{14}}=\sqrt{14-2\sqrt{14}+1}=\sqrt{\left(\sqrt{14}-1\right)^2}=\sqrt{14}-1\)

3, \(\sqrt{21+8\sqrt{5}}=\sqrt{21+2.4\sqrt{5}}=\sqrt{16+2.4\sqrt{5}+5}\)

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

a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)

\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)

\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)

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

\(=\left|7-\sqrt{21}\right|+\left|\sqrt{21}-3\right|\)

\(=7-\sqrt{21}+\sqrt{21}-3\)

\(=4\)

b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)

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

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

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

\(=2\)

Lời giải:
\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{\frac{6-2\sqrt{5}}{2}}+\sqrt{\frac{6+2\sqrt{5}}{2}}\)

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

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

\(B=\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)

\(=\sqrt{18+2\sqrt{18.3}+3}+\sqrt{18-2\sqrt{18.3}+3}\)

\(=\sqrt{(\sqrt{18}+\sqrt{3})^2}+\sqrt{(\sqrt{18}-\sqrt{3})^2}\)

\(=\sqrt{18}+\sqrt{3}+\sqrt{18}-\sqrt{3}=2\sqrt{18}=6\sqrt{2}\)

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\(C=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Rightarrow C^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{(4+\sqrt{10+2\sqrt{5}})(4-\sqrt{10+2\sqrt{5}})}\)

\(8+2\sqrt{4^2-(10+2\sqrt{5})}=8+2\sqrt{6-2\sqrt{5}}\)

\(=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{(\sqrt{5}-1)^2}\)

\(=8+2(\sqrt{5}-1)=6+2\sqrt{5}=(\sqrt{5}+1)^2\)

\(\Rightarrow C=\sqrt{5}+1\)

Lời giải:
\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{\frac{6-2\sqrt{5}}{2}}+\sqrt{\frac{6+2\sqrt{5}}{2}}\)

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

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

\(B=\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)

\(=\sqrt{18+2\sqrt{18.3}+3}+\sqrt{18-2\sqrt{18.3}+3}\)

\(=\sqrt{(\sqrt{18}+\sqrt{3})^2}+\sqrt{(\sqrt{18}-\sqrt{3})^2}\)

\(=\sqrt{18}+\sqrt{3}+\sqrt{18}-\sqrt{3}=2\sqrt{18}=6\sqrt{2}\)

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\(C=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Rightarrow C^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{(4+\sqrt{10+2\sqrt{5}})(4-\sqrt{10+2\sqrt{5}})}\)

\(8+2\sqrt{4^2-(10+2\sqrt{5})}=8+2\sqrt{6-2\sqrt{5}}\)

\(=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{(\sqrt{5}-1)^2}\)

\(=8+2(\sqrt{5}-1)=6+2\sqrt{5}=(\sqrt{5}+1)^2\)

\(\Rightarrow C=\sqrt{5}+1\)

a, \(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=\left(-\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=-1\)

b.\(\sqrt{16+2\sqrt{16.5}+5}+\sqrt{16-2\sqrt{16.5}+5}=\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}=8\)

d,dat \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\Rightarrow A^2=4+\sqrt{7}+2\sqrt{16-7}+4-\sqrt{7}\)\(A^2=8+6=14\Rightarrow A=\sqrt{14}\)

C,\(\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}=\sqrt{17-4\left(2+\sqrt{5}\right)}=\sqrt{17-8-4\sqrt{5}}=\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)