Помогите пожалуйста :(

№4.

1) $2\sqrt{3}+5\sqrt{12}-3\sqrt{27}=2\sqrt{3}+5\sqrt{4}\sqrt{3}-3\sqrt{9}\sqrt{3}==2\sqrt{3}+10\sqrt{3}-9\sqrt{3}=3\sqrt{3}$

$(\sqrt{32}-\sqrt{8})\sqrt{2}=\sqrt{32}\sqrt{2}-\sqrt{8}\sqrt{2}=\sqrt{64}-\sqrt{16}=6-4=2$

$(\sqrt{5}-2)^2=(\sqrt{5})^2+2*\sqrt{5}*2+2^2=5+4+4\sqrt{5}=4\sqrt{5}+9$

$(\sqrt{6}+4\sqrt{3})(\sqrt{6}-4\sqrt{3})=(\sqrt{6})^2-(4\sqrt{3})^2=6-16*3=6-48=-42$

№5

1) $3\sqrt{7}=\sqrt{9}\sqrt{7}=\sqrt{63}\\4\sqrt{6}=\sqrt{16}\sqrt{6}=\sqrt{96}\\63<96;\sqrt{63}<\sqrt{96}\\3\sqrt{7}<4\sqrt{6}$

$5\sqrt{\frac{7}{5}}=\sqrt{25*\frac{7}{5}}=\sqrt{35}\\\frac{1}{2} \sqrt{140}=\sqrt{\frac{140}{4} }=\sqrt{35}\\5\sqrt{\frac{7}{5}}=\frac{1}{2} \sqrt{140}$