Решите пожайлуста за 20 минут срочно

1.a) $4 \sqrt{2}+ \sqrt{50}- \sqrt{18}=4 \sqrt{2}+ \sqrt{25*2}- \sqrt{9*2}= 4 \sqrt{2}+5 \sqrt{2}-3 \sqrt{2}=$ $\sqrt{2}(4+5-3)= 6\sqrt{2} ;$
б) $\sqrt{3}*(2 \sqrt{3}+ \sqrt{12})= \sqrt{3}*(2 \sqrt{3}+\sqrt{4*3})=\sqrt{3}*(2\sqrt{3}+2\sqrt{3})=$ $\sqrt{3}*(\sqrt{3}(2+2))=\sqrt{3}*4\sqrt{3}=4*(\sqrt{3})^2=4*3=12;$
в) $(\sqrt{5}-2)^2=(\sqrt{5})^2-2*2\sqrt{5}+2^2=5-4\sqrt{5}+4=9-4\sqrt{5};$
г) $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=(\sqrt{3})^2-(\sqrt{2})^2=3-2=1;$
2.а) $\frac{3-\sqrt{3} }{2\sqrt{3} }= \frac{\sqrt{3} (3-\sqrt{3) } {\sqrt{3}*2\sqrt{3} }= \frac{3\sqrt{3}-(\sqrt{3})^2 }{2*(\sqrt{3})^2 }= \frac{3\sqrt{3}-3 }{2*3}= \frac{3*(\sqrt{3}-1) }{2*3}= \frac{\sqrt{3}-1 }{2};$
б) $\frac{4b-2}{2\sqrt{b}-\sqrt{2} }= \frac{(\sqrt{4b})^2-(\sqrt{2})^2 }{2\sqrt{b}-\sqrt{2} } = \frac{(\sqrt{4b}-\sqrt{2})(\sqrt{4b}+\sqrt{2})}{2\sqrt{b}-\sqrt{2} }= \frac{(2\sqrt{b}-\sqrt{2})(2\sqrt{b}+\sqrt{2})}{2\sqrt{b}-\sqrt{2} }=$ $2\sqrt{b}+\sqrt{2};$
$3)3 \sqrt{7}<4\sqrt{5}$
$\sqrt{9*7}<\sqrt{16*5}$
$\sqrt{63}\ \textless \ \sqrt{80}$