Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)

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

=1

\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)

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

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)

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

\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)

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

\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)

\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=2\sqrt{5}+4\sqrt{2}\)

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)

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

b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

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

=6

c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)

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

=10

d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)

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

\(=2\sqrt{5}+4\sqrt{2}\)

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

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

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

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

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

\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)

\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)

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

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

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

\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)

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

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)

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

\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)

\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)

\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)

\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)

\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)

a) Ta có: \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

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

\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

b) Ta có: \(\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}\)

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

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

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

c) Ta có: \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

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

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

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

\(=\sqrt{5}\)

d) Ta có: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)

\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)

\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)

\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)

\(=\sqrt{11+6\sqrt{2}}\)

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

a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

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

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

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

b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)

\(\Leftrightarrow B^3+9B-10=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))

c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)

\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)

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

\(\Rightarrow C=1\)

d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

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

\(=\sqrt[3]{3}+\sqrt[3]{2}\)

Vậy...

a, \(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{5-8}}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)

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

\(=\sqrt{10+2\left(\sqrt{5}-2\right)}\)

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

\(=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

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

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

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

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

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

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

\(=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

b) \(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+5}=3\)

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

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

\(=4\sqrt{5}\)