Question

PFV ME AJUDEM!!!!!!!!!

Answer 1

$\frac{\frac{2a^2}{3a+bc} .\sqrt{\frac{1}{b+c} } }{\frac{(b+2b)^2-5c}{18ac} } \\\\Encima:\\\frac{2a^2.\sqrt{1} }{(3a+bc)\sqrt{b+c} } =\frac{2a^2}{(3a+bv)\sqrt{b+c} } \\\\Embaixo:\\\frac{9b^2-5c}{18ac} \\\\Juntando:\\\\\frac{(18ac)(2a^2)}{(9b^2-5c)[(3a+bc)\sqrt{b+c}]} \\\\\frac{36a^3c}{(9b^2-5c)(3a\sqrt{b+c}+bc\sqrt{b+c}) } \\\\\frac{36a^3c}{(27ab^2\sqrt{b+c})+(9b^3c\sqrt{b+c})-(15ac\sqrt{b+c})-(5b^2c\sqrt{b+c}) }\\\\\frac{36a^3c}{(27ab^2+9b^3c-15ac-5b^2c)\sqrt{b+c} }$

É o máximo que se pode ir sem os valores.