Question

√96(√162-√288-+√128-√32)

Answer 1

$\sqrt{96}=4\sqrt{6}$

$\sqrt{162} =9\sqrt{2}$

$\sqrt{288}=12\sqrt{2}$

$\sqrt{128}=8\sqrt{2}$

$\sqrt{32} =4\sqrt{2}$

$4\sqrt{6} (9\sqrt{2} -12\sqrt{2} +8\sqrt{2} -4\sqrt{2} )$

$4\sqrt{6} *\sqrt{2}$

$4\sqrt{12} =8\sqrt{3}$