[tex]\it \dfrac{a}{a-2}=\dfrac{b+15}{b+5} \stackrel{derivare}{\Longrightarrow}\ \dfrac{a}{a-2-a}=\dfrac{b+15}{b+5-b-15}\Rightarrow \dfrac{a}{-2}=\dfrac{b+15}{-10}|_{\cdot(-2)}\Rightarrow\\ \\ \\ \Rightarrow a=\dfrac{b+15}{5}\ \ \ \ \ (*)\\ \\ \\ (*) \Rightarrow \dfrac{a}{b}=\dfrac{\dfrac{b+15}{5}}{b}=\dfrac{b+15}{5b}=\dfrac{\ b^{(b}}{5b}+\dfrac{\ 15^{(5}}{5b}=\dfrac{1}{5}+\dfrac{3}{b} \Rightarrow\\ \\ \\ \Rightarrow max(\dfrac{a}{b}) =max(\dfrac{3}{b}) =min(b) \Rightarrow b=1[/tex]
[tex]\it b=1 \Rightarrow \dfrac{a}{b}-\dfrac{a}{1}=a\ \stackrel{(8)}{=}\ \dfrac{1+15}{5}=\dfrac{^{2)}16}{\ 5}=\dfrac{32}{10}=3,2\\ \\ \\ Prin\ urmare,\ max(\dfrac{a}{b})=3,2[/tex]
