Второй вариант, пожалуйста, помогите решить

буду по одному добавлять [latex]1) 3 \sqrt{5} = \sqrt{3^2*5}= \sqrt{45} \\ -5 \sqrt{1 \frac{3}{5} } = \sqrt{(-5)^2* \frac{8}{5} }= \sqrt{ \frac{200}{5} }= \sqrt{40} \\ \frac{1}{4} \sqrt{48} = \sqrt{( \frac{1}{4})^2*48 }= \sqrt{ \frac{48}{16} } = \sqrt{3} [/latex] [latex]2) =2*3 \sqrt{3}+4*4 \sqrt{3} - \frac{1}{5} *5 \sqrt{3}-9 \sqrt{3}=6 \sqrt{3}+16\sqrt{3}-\sqrt{3}-9\sqrt{3} \\ =12\sqrt{3} [/latex] [latex]4) =24+21 \sqrt{2} -8\sqrt{2}-14-13\sqrt{2}=10; \\ =(3\sqrt{2}+2)(3\sqrt{2}+2)+(6-\sqrt{2})(6-\sqrt{2})=18+12\sqrt{2}+4+12- \\ 12\sqrt{2}+2=36; \\ ( \sqrt{11}-2\sqrt{2})( \sqrt{11}+2\sqrt{2})+20=11-8+20=23 [/latex] [latex]5) =\frac{15* \sqrt{6} }{2\sqrt{6}*\sqrt{6}}= \frac{15\sqrt{6}}{12} = \frac{5\sqrt{6}}{4}; \\ = \frac{19*(2 \sqrt{5}+1) }{(2\sqrt{5} -1)(2\sqrt{5} +1)} = \frac{38\sqrt{5} +19}{20-1} = \frac{19(2\sqrt{5} +1)}{19}=2\sqrt{5} +1 [/latex] [latex]6) 8 \sqrt{ \frac{3}{4} } \ \textgreater \ \frac{1}{3} \sqrt{405} \\ \sqrt{8^2* \frac{3}{4} } \ \textgreater \ \sqrt{( \frac{1}{3})^2*405 } \\ \sqrt{ \frac{192}{4} } \ \textgreater \ \sqrt{ \frac{405}{9} } \\ \sqrt{48} \ \textgreater \ \sqrt{45} [/latex]