Question

Sa se afle valoarea expresiei $\frac{a+b}{a-b}$, stiind ca 2a^2+2b^2=5ab si b>a>0

Answer 1

[tex]\it 2a^2+2b^2=5ab|_{+4ab} \Rightarrow 2a^2+2b^2 + 4ab =9ab \Rightarrow \\\;\\ \Rightarrow \it2(\it a^2+b^2+\it2ab) =9ab \Rightarrow 2(a+b)^2 =9ab \ \ \ (1)[/tex]

$\it 2a^2+2b^2=5ab|_{-4ab} \Rightarrow 2a^2+2b^2 - 4ab =ab \Rightarrow \\\;\\ \Rightarrow \it2(\it a^2+b^2-\it2ab) = ab \Rightarrow 2(a-b)^2 =ab \ \ \ (2)$

$\it (1),\ (2) \Longrightarrow \dfrac{2(a+b)^2}{2(a-b)^2} = \dfrac{9ab}{ab} \Longrightarrow \dfrac{(a+b)^2}{(a-b)^2}= 9 \Longrightarrow$


$\it \Longrightarrow \sqrt{ \dfrac{(a+b)^2}{(a-b)^2}} = \sqrt9 \Longrightarrow \dfrac{|a+b|}{|a-b|} =3$

Deoarece b > a > 0, ultima egalitate devine:

$\it \dfrac{a+b}{-(a-b)} =3|_{\cdot(-1)} \Longrightarrow \dfrac{a+b}{a-b} = -3.$