Question

ПОМОГИТЕ РЕШИТЬ!
Даю много баллов! ​

Answer 1

$1)(\sqrt{11}-\sqrt{3})(\sqrt{11}+\sqrt{3})=(\sqrt{11})^{2}-(\sqrt{3} )^{2}=11-3=8\\\\2)(\sqrt{24}-\sqrt{6})*\sqrt{6}=(\sqrt{4*6}-\sqrt{6})*\sqrt{6}=(2\sqrt{6}-\sqrt{6})*\sqrt{6}=\sqrt{6}*\sqrt{6}=6\\\\3)\frac{39}{(2\sqrt{13})^{2}}=\frac{39}{2^{2}*(\sqrt{13})^{2}} =\frac{39}{4*13}=\frac{3}{4}=0,75$

$4)\frac{4}{5}\sqrt{5}*\sqrt{45}=\frac{4}{5}\sqrt{5}*\sqrt{9*5}=\frac{4}{5}\sqrt{5}*3\sqrt{5} =\frac{4*3*5}{5} =12\\\\5)\frac{\sqrt{162} }{\sqrt{2}}=\sqrt{\frac{162}{2}}=\sqrt{81}=\sqrt{9^{2}}=9\\\\6)\sqrt{10}*\sqrt{2,5}=\sqrt{10*2,5}=\sqrt{25}=\sqrt{5^{2}}=5$

$6)\frac{\sqrt{288} }{3\sqrt{8}}=\frac{\sqrt{144*2}}{3*\sqrt{4*2}}=\frac{12\sqrt{2}}{3*2\sqrt{2}}=\frac{12\sqrt{2}}{6\sqrt{2}}=2\\\\7)\sqrt{4^{2}*3^{4}} =\sqrt{4^{2}*(3^{2})^{2}}=4*3^{2}=4*9=36\\\\8)\sqrt{610^{2}-448^{2}} =\sqrt{(610-448)(610+448)}=\sqrt{162*1058}=\sqrt{81*2*529*2}=\sqrt{9^{2}*2^{2}*23^{2}} =9*2*23=414$