Question

a^3+b^3+3b^2+3b+1/a^2-ab-a+(b+1)^2 при а=4 - корень из 3, b=4+ корень из 3

Answer 1

$asqrt{3}+bsqrt{3}+3bsqrt{2}+3b+frac{1}{a}sqrt{2}-ab-a+(b+1)sqrt{2}$

$sqrt{3}(a+b)+sqrt{2}(3b+frac{1}{a}+b+1)+3b-ab-a$

1) sqrt{3}(4-sqrt{3}+4+sqrt{3})=4sqrt{3}\2) sqrt{2}(34+sqrt{3}+frac{1}{4-sqrt{3}}+4+sqrt{3}+1)=sqrt{2}(17+4sqrt{4}+frac{1}{4-sqrt{3}})=\=17sqrt{2}+4sqrt{6}+frac{sqrt{2}}{4+sqrt{3}}\3) 12+3sqrt{3}-16+3-4+sqrt{3}=4sqrt{3}-5\4) $1) sqrt{3}(4-sqrt{3}+4+sqrt{3})=4sqrt{3}\2) sqrt{2}(3b+frac{1}{4-sqrt{3}}+4+sqrt{3}+1)=sqrt{2}(17+4sqrt{4}+frac{1}{4-sqrt{3}})=\=17sqrt{2}+4sqrt{6}+frac{sqrt{2}}{4-sqrt{3}}\3) 12+3sqrt{3}-16+3-4+sqrt{3}=4sqrt{3}-5\4) 4sqrt{3}+4sqrt{3}-5+17sqrt{2}+4sqrt{6}+frac{sqrt{2}}{4-sqrt{3}}=\=8sqrt{3}+17sqrt{2}+frac{sqrt{2}}{4-sqrt{3}}=$$frac{32sqrt{3}-24+68sqrt{2}-17sqrt{6}+sqrt{2}}{4-sqrt{3}}=\=frac{69sqrt{2}+32sqrt{3}-17sqrt{6}-24}{4-sqrt{3}}$

 дальше я не знаю как делать...((