Question

Упрос$(3\sqrt{6} +2\sqrt8-\sqrt32)*\sqrt2-\sqrt108$тить

Answer 1

$(3 \sqrt{6} +2 \sqrt{8} - \sqrt{32} )* \sqrt{2} - \sqrt{108} =\\ =(3 \sqrt{6} +4 \sqrt{2} - 4\sqrt{2} )* \sqrt{2} - 6\sqrt{3} = \\ =3 \sqrt{6} * \sqrt{2} - 6\sqrt{3} = 3 \sqrt{12} -6\sqrt{3} =6\sqrt{3} -6\sqrt{3} =0$