Question

media geometrica a numerelor a si b pentru care:Rad(9a2-12a+5) +rad(4b2-12b+13)<=3 este 1/2 sau 1/3 sau 1/6 sau 1

Answer 1

Răspuns:

Explicație pas cu pas:

$\sqrt{9a^2-12a+5}+\sqrt{4b^2-12b+13}\leq 3\\\sqrt{9a^2-12a+4+1}+\sqrt{4b^2-12b+9+4}\leq 3\\\sqrt{(3a-2)^2+1} +\sqrt{(2b-3)^2+4}\leq 3\\\texttt{Observam ca:}\\\sqrt{(3a-2)^2+1} \geq 1 \texttt{ si }\sqrt{(2b-3)^2+4}\geq \sqrt{4} = 2\\\texttt{Egalitatea are loc atunci cand: }\\\\3a-2 = 0\Rightarrow a = \dfrac{2}{3}\\2b-3 = 0\Rightarrow b = \dfrac{3}{2}\\m_g=\sqrt{a\cdot b}=\sqrt{\dfrac{2}{3}\cdot \dfrac{3}{2}} = \boxed{1}$