Решите пожайлуста за 20 минут срочно

1.a)[latex]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}= [/latex] [latex] sqrt{2}(4+5-3)= 6sqrt{2} ;[/latex]б)[latex] sqrt{3}*(2 sqrt{3}+ sqrt{12})= sqrt{3}*(2 sqrt{3}+sqrt{4*3})=sqrt{3}*(2sqrt{3}+2sqrt{3})= [/latex] [latex]sqrt{3}*(sqrt{3}(2+2))=sqrt{3}*4sqrt{3}=4*(sqrt{3})^2=4*3=12;[/latex]в)[latex](sqrt{5}-2)^2=(sqrt{5})^2-2*2sqrt{5}+2^2=5-4sqrt{5}+4=9-4sqrt{5};[/latex]г)[latex](sqrt{3}-sqrt{2})(sqrt{3}+sqrt{2})=(sqrt{3})^2-(sqrt{2})^2=3-2=1; [/latex]2.а)[latex] frac{3-sqrt{3} }{2sqrt{3} }= frac{sqrt{3} (3-sqrt{3) } {sqrt{3}*2sqrt{3} }= frac{3sqrt{3}-(sqrt{3})^2 }{2*(sqrt{3})^2 }= frac{3sqrt{3}-3 }{2*3}= frac{3*(sqrt{3}-1) }{2*3}= frac{sqrt{3}-1 }{2}; [/latex]б)[latex] frac{4b-2}{2sqrt{b}-sqrt{2} }= frac{(sqrt{4b})^2-(sqrt{2})^2 }{2sqrt{b}-sqrt{2} } = frac{(sqrt{4b}-sqrt{2})(sqrt{4b}+sqrt{2})}{2sqrt{b}-sqrt{2} }= frac{(2sqrt{b}-sqrt{2})(2sqrt{b}+sqrt{2})}{2sqrt{b}-sqrt{2} }= [/latex] [latex]2sqrt{b}+sqrt{2};[/latex][latex]3)3 sqrt{7}<4sqrt{5}[/latex][latex]sqrt{9*7}