Question

решите неравенство (картинка внизу)

Answer 1

$frac{a^3+b^3+3b^2+3b+1}{a^2-ab-a+(b+1)^2}=\\frac{a^3+(b+1)^3}{a^2-a(b+1)+(b+1)^2}=\\frac{(a+(b+1))(a^2-a(b+1)+(b+1)^2}{a^2-a(b+1)+(b+1)^2}=a+(b+1)=a+b+1=\\4-sqrt{3}+4+sqrt{3}+1=9$