Question

Решите пожалуйста))) ​

Answer 1

$1)(5\sqrt{18}+\sqrt{18})*\sqrt{2}= 6\sqrt{18}*\sqrt{2}=6\sqrt{18*2}=6\sqrt{36}=6*6=36$

$2)(\sqrt{27}-\sqrt{12}):\sqrt{3}=(\sqrt{27:3}-\sqrt{12:3})=\sqrt{9}-\sqrt{4}=3-2=1$

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

$4)(\sqrt{m}+3)(\sqrt{m}-3)=(\sqrt{m})^2-3^2=m-9$

$5)(\sqrt{13}-\sqrt{5})^2=(\sqrt{13})^2-2*\sqrt{13}*\sqrt{5}+(\sqrt{5})^2=\\\\=13-2\sqrt{65}+5=18-2\sqrt{65}=2*(9-\sqrt{65})$

$6)(5\sqrt{2}+\sqrt{6})^2=(25*\sqrt{2})^2+2*5\sqrt{2}*\sqrt{6}+(\sqrt{6})^2=\\\\=50+10\sqrt{12}+6=56+10*2\sqrt{3}=56+20\sqrt{3}=4*(14+5\sqrt{3})$

$7)(\sqrt{3}+1)(3-\sqrt{3}+1)=(\sqrt{3}+1)(4-\sqrt{3})=\\\\=4\sqrt{3}+4-(\sqrt{3})^2-\sqrt{3}=3\sqrt{3}+4-3=3\sqrt{3}+1$