Question

va rog ajutati-mă a) și b)... ma abonez...dau si coroană​

Answer 1

Aducem numerele a si intr-o forma simplificata si apoi calculam.

a)

$a=\sqrt{3} (\sqrt{27}+\sqrt{75}-\sqrt{243})+2\sqrt{3}*\sqrt{48}\\a=\sqrt{3} (3\sqrt{3}+5\sqrt{3}-9\sqrt{3})+2\sqrt{3}*4\sqrt{3}\\a=\sqrt{3} (-1\sqrt{3})+24\\a=-3+24\\a=21\\\\b=\sqrt{2}(\sqrt{32}+\sqrt{50}-\sqrt{98})-3\sqrt{3}*\sqrt{12}\\b=\sqrt{2}(4\sqrt{2}+5\sqrt{2}-7\sqrt{2})-3\sqrt{3}*2\sqrt{3}\\b=\sqrt{2}(2\sqrt{2})-18\\b=4-18\\b=-14\\\\a+b = 21-14 = 7\\a-b=21+14=35$

b)

$a=\sqrt{8}(\sqrt{108}+\sqrt{243}-\sqrt{192}-\sqrt{75})\\a=2\sqrt{2}(6\sqrt{3}+9\sqrt{3}-8\sqrt{3}-5\sqrt{3})\\a=2\sqrt{2}*2\sqrt{3}\\a=4\sqrt{6}\\\\b=\sqrt{12}(\sqrt{128}-\sqrt{72}-\sqrt{98}+\sqrt{162})\\b=2\sqrt{3}(8\sqrt{2}-6\sqrt{2}-7\sqrt{2}+9\sqrt{2})\\b=2\sqrt{3}*4\sqrt{2}\\b=8\sqrt{6}\\\\a+b=4\sqrt{6}+8\sqrt{6} = 12\sqrt{6}\\a-b=4\sqrt{6}-8\sqrt{6} = -4\sqrt{6}$