Question

[Dương Hoàng Nam]

giải giúp em vs ạ em cảm ơn nhiều ạ.

Answer 1

[Nguyễn Lê Phước Thịnh]

2: Ta có: \(\sqrt{16-6\sqrt{7}}\cdot\left(3+\sqrt{7}\right)\)

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

=9-7

=2

3: Ta có: \(\left(\sqrt{6}+\sqrt{14}\right)\cdot\sqrt{5-2\sqrt{21}}\)

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

=7-3

=4

Answer 2

[Nguyễn Hoàng Minh]

\(1,=\sqrt{\left(5+2\sqrt{6}\right)^2}-\sqrt{\left(3-\sqrt{6}\right)^2}=5+2\sqrt{6}-3+\sqrt{6}=2+3\sqrt{6}\\ 2,=\sqrt{\left(3-\sqrt{7}\right)^2}\left(3+\sqrt{7}\right)=\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)=9-7=2\\ 3,=\left(\sqrt{3}+\sqrt{7}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{3}+\sqrt{7}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\\ =\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=7-3=4\\ 4,=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{6+2\sqrt{5}}+\sqrt{4-2\sqrt{3}}\right)\\ =\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+1+\sqrt{3}-1\right)\\ =\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)

\(5,\\ =\sqrt{\left(3\sqrt{3}-5\right)^2}+\sqrt{\left(5-2\sqrt{3}\right)^2}=3\sqrt{3}-5+5-2\sqrt{3}=\sqrt{3}\\ 6,=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\\ =\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}-2\sqrt{2}\right)^2}\\ =2\sqrt{2}-\sqrt{5}-3\sqrt{5}+2\sqrt{2}=4\sqrt{2}-4\sqrt{5}\)