Question

Найдите значение выражения. $(\sqrt{2}-2\sqrt{3})^{2} + 4\sqrt{6}$

Answer 1

$(\sqrt{2}-2\sqrt{3})^{2}+4\sqrt{6}=(\sqrt{2})^{2} -2*\sqrt{2} *2\sqrt{3}+(2\sqrt{3})^{2}+4\sqrt{6}=\\\\=2-4\sqrt{6}+12+4\sqrt{6} =\boxed{14}$

Answer 2

Ответ:

$14$

Объяснение:

$(\sqrt{2}-2\sqrt{3} )^{2} +4\sqrt{6} = (\sqrt{2} )^{2} -4\sqrt{2} \sqrt{3} +4(\sqrt{3} )^{2} + 4\sqrt{6} =2 -4\sqrt{2} \sqrt{3}+ 4(\sqrt{3} )^{2} + 4\sqrt{6} = 2 - 4\sqrt{6} + 4(\sqrt{3} )^{2} +4\sqrt{6} = 2 - 4\sqrt{6} + 4*3 +4\sqrt{6} = 2 - 4\sqrt{6} + 12 +4\sqrt{6} = 14 - 4\sqrt{6} + 4\sqrt{6} = 14$