Question

Помогите решить пожалуйста. Буду очень признательна очень умному человеку.

Answer 1

Вариант 2

$1)\frac{(6\sqrt{2})^2}{36}=\frac{6^2*2}{36}=\frac{36*2}{36}=2\\\\2)\sqrt{\frac{9}{16}}-\sqrt{\frac{25}{36}}=\sqrt{\frac{3^2}{4^2}}-\sqrt{\frac{5^2}{6^2}}=\frac{3}{4}-\frac{5}{6}=\frac{9}{12}-\frac{10}{12}=-\frac{1}{12}\\\\3)(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{5}*\sqrt{5}-\sqrt{5}*\sqrt{3}+\sqrt{3}*\sqrt{5}-\sqrt{3}*\sqrt{3}=5-3=2$

$4)(\sqrt{12}-\sqrt{3})^2=(\sqrt{12}-\sqrt{3})*(\sqrt{12}-\sqrt{3})=\\\\=12-\sqrt{12}*\sqrt{3}-\sqrt{3}*\sqrt{12}+3=15-2*(\sqrt{3}*\sqrt{12})=\\\\=15-2*\sqrt{3*12}=15-2*\sqrt{36}=15-2*\sqrt{6^2}=15-2*6=15-12=3.\\\\5)0,5\sqrt{0,04}-\frac{1}{6}\sqrt{144}=0,5\sqrt{0,2*0,2}-\frac{1}{6}*\sqrt{12*12}=\\\\=0,5*0,2-\frac{1}{6}*12=0,1-2=-1,9$

$6)10\sqrt{3}-4\sqrt{48}+\sqrt{27}=10\sqrt{3}-4\sqrt{16*3}+\sqrt{9*3}=\\\\=10\sqrt{3}-4\sqrt{4^2*3}+\sqrt{3^2*3}=10\sqrt{3}-4*4\sqrt{3}+3\sqrt{3}=\\\\=(10-16+3)*\sqrt{3}=-3\sqrt{3}$

$7)\frac{\sqrt{27}}{\sqrt{3}}=\frac{\sqrt{9*3}}{\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{3}}=3\\\\8)\sqrt{3^4*2^6}=\sqrt{(3^2)^2*(2^3)^2}=3^2*2^3=3*3*2*2*2=9*8=72\\.$

Вариант 3

$1)\frac{(3\sqrt{6})^2}{12}=\frac{3*3*6}{12}=\frac{3*3*3*2}{3*2*2}=\frac{3*3}{2}=\frac{9}{2}=4\frac{1}{2}\\\\2)\sqrt{\frac{1}{16}}+\sqrt{\frac{25}{36}}=\sqrt{\frac{1*1}{4*4}}+\sqrt{\frac{5*5}{6*6}}=\frac{1}{4}+\frac{5}{6}=\frac{3}{12}+\frac{10}{12}=\frac{13}{12}=1\frac{1}{12}\\\\3)(\sqrt{12}+\sqrt{3})(\sqrt{12}-\sqrt{3})=12-3=9$

$4)(\sqrt{2}+\sqrt{3})=2+2\sqrt{2*3}+3=5+2\sqrt{6}\\\\5)0,2\sqrt{0,04}+\frac{1}{2}\sqrt{144}=0,2*0,2+\frac{1}{2}*12=0,04+6=6,04\\\\6)2\sqrt{3}-4\sqrt{48}+\sqrt{12}=2\sqrt{3}-4*\sqrt{16*3}+\sqrt{4*3}=\\\\=2\sqrt{3}-4*4\sqrt{3}+2\sqrt{3}=-16\sqrt{3}$

$7)\sqrt{2}*\sqrt{18}=\sqrt{2*18}=\sqrt{36}=6\\\\8)\sqrt{2^2*5^6}=2\sqrt{5^6}=2\sqrt{(5^3)^2}=2*5^3=2*5*5*5=250.$