Question

Даю все свои баллы пажалуйста ​

Answer 1

№1

$\displaystyle\\\sqrt{36b}-\sqrt{16b} +2\sqrt{b} =6\sqrt{b}-4\sqrt{b}+2\sqrt{b}=2\sqrt{b}+2\sqrt{b} =4\sqrt{b} \\\\\sqrt{81b}-\sqrt{25b}+3\sqrt{b}=9\sqrt{b}-5\sqrt{b}+3\sqrt{b}=4\sqrt{b}+3\sqrt{b}=7\sqrt{b}$

№2

$\displaystyle\\(3\sqrt{8}+\sqrt{18})*\sqrt{2} =(\sqrt{72}+\sqrt{18})*\sqrt{2}= \sqrt{72}*\sqrt{2}+\sqrt{18}*\sqrt{2}=\\\\=\sqrt{144} +\sqrt{36}=12+6=18\\\\\\(2a-\sqrt{b})(2a+\sqrt{b})=2^2a^2-\sqrt{b^2}=4a^2-b\\\\\\(\sqrt{3}+\sqrt{2})^2-\sqrt{24}=\sqrt{3^3}+2*\sqrt{3}*\sqrt{2}+\sqrt{2^2}-\sqrt{24} =3+2\sqrt{6}+2-\sqrt{24}=\\\\=3+\sqrt{24}+2-\sqrt{24}=3+2=5$

$\sqrt{3} *(2\sqrt{12}+\sqrt{48}) =\sqrt{3} *(\sqrt{48}+\sqrt{48})=\sqrt{3}*\sqrt{48}+\sqrt{3}*\sqrt{48}=\\\\=\sqrt{144}+\sqrt{144}=12+12=24\\\\\\(\sqrt{a}+3b)(\sqrt{a}-3b)=\sqrt{a^2}-3^2b^2=a-9b^2\\\\\\(\sqrt{5}-\sqrt{2})^2+\sqrt{40}=\sqrt{5^2}-2*\sqrt{5}*\sqrt{2}+\sqrt{2^2}+\sqrt{40}=5-2\sqrt{10 }+2+\sqrt{40}=\\\\=5-\sqrt{40}+2+\sqrt{40}=5+2=7$