Question

Ma poate ajuta cineva ,va rog la exercițiul 14, este urgent,dau coroană

Answer 1

Răspuns:

$\frac{2}{3}<\frac{3}{a}<\frac{7}{5}<\frac{3}{b}<\frac{6a}{2a+3b}$

a+b=?

a, b ∈ N*

$\frac{2}{3}<\frac{3}{a}$ ⇒ 2a<9 ⇒ $a<\frac{9}{2}$

$\frac{3}{a}<\frac{7}{5}$ ⇒ 7a>15 ⇒ $a>\frac{15}{7}$

$\frac{15}{7}<a<\frac{9}{2}$ ⇒ a∈{3, 4}

$\frac{7}{5}<\frac{3}{b}$ ⇒ 7b<15 ⇒ $b<\frac{15}{7}$ ⇒ b∈{1, 2}

daca b=1, a=3

        $\frac{2}{3}<\frac{3}{3}<\frac{7}{5}<\frac{3}{1}<\frac{6*3}{2*3+3*1}=\frac{18}{9}=2$   FALS

daca b=1, a=4

        $\frac{2}{3}<\frac{3}{4}<\frac{7}{5}<\frac{3}{1}<\frac{6*4}{2*4+3*1}=\frac{24}{11}=2\frac{2}{11}$   FALS      

daca b=2, a=3

        $\frac{2}{3}<\frac{3}{3}<\frac{7}{5}<\frac{3}{2}<\frac{6*3}{2*3+3*2}=\frac{18}{12}=\frac{3}{2}$   FALS    

daca b=2, a=4

        $\frac{2}{3}<\frac{3}{4}<\frac{7}{5}<\frac{3}{2}<\frac{6*4}{2*4+3*2}=\frac{24}{14}=1\frac{10}{14}$   ADEVARAT

a+b=4+2=3    

Explicație pas cu pas: