Question

rad 27×(rad32-rad72+rad162-rad288)-rad8(2rad108-5rad192)​

Answer 1

Răspuns:

41√6

Explicație pas cu pas:

$\sqrt{27} (\sqrt{32} -\sqrt{72} +\sqrt{162} -\sqrt{288} )-\sqrt{8} (2\sqrt{108} -5\sqrt{192} )=$

$=\sqrt{9*3} (\sqrt{16*2} -\sqrt{36*2} +\sqrt{81*2} -\sqrt{144*2} )-\sqrt{4*2} (2\sqrt{36*3} -5\sqrt{64*3} )=$

$=3\sqrt{3} (4\sqrt{2} -6\sqrt{2} +9\sqrt{2} -12\sqrt{2} )-2\sqrt{2} (12\sqrt{3} -40\sqrt{3} )=$

$=3\sqrt{3} (-5\sqrt{2} )-2\sqrt{2} (-28\sqrt{3} )=$

$=-15\sqrt{6}+56\sqrt{6} =$

$=41\sqrt{6}$