Question

Пожалуйста!даю много балловвв!

Answer 1

$frac{sqrt{18}}{3}+3sqrt{8}-sqrt{98} = \ sqrt{frac{18}{9}}+sqrt{9*8}-sqrt{98} = \ sqrt{2}+6sqrt{2}-sqrt{98}=\ 7sqrt{2}-sqrt{98}=sqrt{98}-sqrt{98}=0\ \ 2sqrt{5}(sqrt{20}-3sqrt{5})\ 2*10-6*5=-10\ \ (3+2sqrt{7})^2=9+12sqrt{7}+28\ 12sqrt{7}+37\ frac{9a-b^2}{9a-6bsqrt{a}+b^2}=frac{(3sqrt{a}-b)(3*sqrt{a}+b)}{(3sqrt{a}+b)^2} = frac{ 3sqrt{a}-b}{3sqrt{a}+b}$