Question

Tính:

a)\(\sqrt{3\sqrt{2}-2\sqrt{3}}\cdot\sqrt{3\sqrt{2}+2\sqrt{3}}\)

b) \(\sqrt{2+2\sqrt{2-\sqrt{2}}}\cdot\sqrt{2-2\sqrt{2-\sqrt{2}}}\)

c)\(\left(\sqrt{2}-\sqrt{7}\right)\sqrt{9+2\sqrt{14}}\)

Answer 1

a)\(\sqrt{3\sqrt{2}-2\sqrt{3}}.\sqrt{3\sqrt{2}+2\sqrt{3}}\)

= \(\sqrt{18-12}\)

= \(\sqrt{6}\)

b) \(\sqrt{2+2\sqrt{2-\sqrt{2}}}.\sqrt{2-2\sqrt{2-\sqrt{2}}}\)

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

= \(\sqrt{4-4.\left(2-4\sqrt{2}+2\right)}\)

= \(\sqrt{4-8+16\sqrt{2}-8}\)

= \(\sqrt{-12+16\sqrt{2}}\)

c) 

\(\left(\sqrt{2}-\sqrt{7}\right).\sqrt{9+2\sqrt{14}}\)

= \(\left(\sqrt{2}-\sqrt{7}\right).\left(2+2\sqrt{7}.\sqrt{2}+7\right)\)

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

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

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