1: \(\frac{\sqrt{5+2\sqrt6}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
\(=\frac{\sqrt{\left(\sqrt3+\sqrt2\right)^2}+\sqrt{\left(\sqrt5-\sqrt3\right)^2}}{\sqrt{\left(\sqrt5+\sqrt2\right)^2}}\)
\(=\frac{\sqrt3+\sqrt2+\sqrt5-\sqrt3}{\sqrt5+\sqrt2}=\frac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}\)
=1
2: \(\frac{3}{2+\sqrt3}+\frac{13}{4-\sqrt3}+\frac{6}{\sqrt3}\)
\(=\frac{3\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}+\frac{13\left(4+\sqrt3\right)}{\left(4-\sqrt3\right)\left(4+\sqrt3\right)}+2\sqrt3\)
\(=3\left(2-\sqrt3\right)+4+\sqrt3+2\sqrt3=6-3\sqrt3+3\sqrt3+4=10\)
3: \(\frac{3}{2+\sqrt3}+\frac{13}{4-\sqrt3}+\frac{6}{\sqrt3}\)
\(=\frac{3\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}+\frac{13\left(4+\sqrt3\right)}{\left(4-\sqrt3\right)\left(4+\sqrt3\right)}+2\sqrt3\)
\(=3\left(2-\sqrt3\right)+4+\sqrt3+2\sqrt3=6-3\sqrt3+3\sqrt3+4=10\)
4: \(\sqrt{\frac{2-\sqrt3}{2+\sqrt3}}+\sqrt{\frac{2+\sqrt3}{2-\sqrt3}}\)
\(=\sqrt{\frac{\left(2-\sqrt3\right)^2}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}}+\sqrt{\frac{\left(2+\sqrt3\right)^2}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}}\)
\(=\sqrt{\left(2+\sqrt3\right)^2}+\sqrt{\left(2-\sqrt3\right)^2}\)
\(=2+\sqrt3+2-\sqrt3\)
=4
5: \(3\sqrt2-\sqrt{13+30\cdot\sqrt{2+\sqrt{9+4\sqrt2}}}\)
\(=3\sqrt2-\sqrt{13+30\cdot\sqrt{2+\sqrt{\left(2\sqrt2+1\right)^2}}}\)
\(=3\sqrt2-\sqrt{13+30\cdot\sqrt{2+2\sqrt2+1^{}}}\)
\(=3\sqrt2-\sqrt{13+30\cdot\sqrt{\left(\sqrt2+1\right)^2}}\)
\(=3\sqrt2-\sqrt{13+30\cdot\left(\sqrt2+1\right)^{}}\)
\(=3\sqrt2-\sqrt{43+30\sqrt2}\)
\(=3\sqrt2-\sqrt{25+2\cdot5\cdot3\sqrt2+18}\)
\(=3\sqrt2-\sqrt{\left(5+3\sqrt2\right)^2}=3\sqrt2-5-3\sqrt2=-5\)
6: \(\left(\frac{3\sqrt5+5\sqrt3}{\sqrt3+\sqrt5}+1\right)\left(1-\frac{3\sqrt5-\sqrt{15}}{\sqrt3-1}\right)\)
\(=\left(\frac{\sqrt{15}\left(\sqrt5+\sqrt3\right)}{\left(\sqrt5+\sqrt3\right)}+1\right)\left(1-\frac{\sqrt{15}\left(\sqrt3-1\right)}{\sqrt3-1}\right)\)
\(=\left(1+\sqrt{15}\right)\left(1-\sqrt{15}\right)=1-15=-14\)
7: Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)
=>\(A^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt5}\right)\left(4-\sqrt{10+2\sqrt5}\right)}\)
=>\(A^2=8+2\cdot\sqrt{16-10-2\sqrt5}=8+2\cdot\sqrt{6-2\sqrt5}\)
=>\(A^2=8+2\cdot\sqrt{\left(\sqrt5-1\right)^2}=6+2\sqrt5\)
=>\(A^2=\left(\sqrt5+1\right)^2\)
=>\(A=\sqrt5+1\)
8: \(\left(\frac{2}{\sqrt3-1}+\frac{3}{\sqrt3-2}+\frac{15}{3-\sqrt3}\right)\cdot\frac{1}{5+\sqrt3}\)
\(=\left(\frac{2\left(\sqrt3+1\right)}{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}-\frac{3\left(2+\sqrt3\right)}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}+\frac{15\left(3+\sqrt3\right)}{\left(3-\sqrt3\right)\left(3+\sqrt3\right)}\right)\cdot\frac{1}{5+\sqrt3}\)
\(=\left\lbrack\sqrt3+1-3\left(2+\sqrt3\right)+\frac{15\left(3+\sqrt3\right)}{9-3}\right\rbrack\cdot\frac{1}{5+\sqrt3}\)
\(=\left(-2\sqrt3-5+2,5\left(3+\sqrt3\right)\right)\cdot\frac{1}{5+\sqrt3}=\left(-2\sqrt3-5+7,5+2,5\sqrt3\right)\cdot\frac{1}{5+\sqrt3}\)
\(=\frac{2,5+0,5\cdot\sqrt3}{5+\sqrt3}=\frac{0,5\left(5+\sqrt3\right)}{5+\sqrt3}=0,5\)
10: \(\frac{1}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{1}{\sqrt2-\sqrt{2-\sqrt3}}\)
\(=\frac{\sqrt2}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2}{2-\sqrt{4-2\sqrt3}}\)
\(=\frac{\sqrt2}{2+\sqrt3+1}+\frac{\sqrt2}{2-\sqrt3+1}=\frac{\sqrt2}{3+\sqrt3}+\frac{\sqrt2}{3-\sqrt3}\)
\(=\frac{\sqrt2\left(3-\sqrt3\right)+\sqrt2\left(3+\sqrt3\right)}{6}=\frac{6\sqrt2}{6}=\sqrt2\)
