tâm là giao của 2 đường chéo AC và BD

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

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

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

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

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

a, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9}{4}-\frac{4\sqrt{2}}{4}}=\sqrt{\frac{9-4\sqrt{2}}{4}}\)

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

b, \(\sqrt{\frac{129}{16}+\sqrt{2}}=\sqrt{\frac{129+16\sqrt{2}}{16}}=\sqrt{\frac{\left(8\sqrt{2}\right)^2+16\sqrt{2}+1}{16}}\)

\(=\sqrt{\frac{\left(8\sqrt{2}+1\right)^2}{16}}=\frac{8\sqrt{2}+1}{4}\)

xin loi chi em moi lop 5

Bài 4 : 

a, \(\sqrt{\left(4-3\sqrt{2}\right)^2}=4-3\sqrt{2}\)

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

e, \(\frac{\sqrt{2}-\sqrt{11+6\sqrt{2}}}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}=\frac{\sqrt{2}+\sqrt{3^2+2.3\sqrt{2}+2}}{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}}\)

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

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

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