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Levisfans
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Am nevoie de ajutor la subiectul III . Va rog
DAU COROANA


Am Nevoie De Ajutor La Subiectul III Va Rog DAU COROANA class=

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[tex]\it 3.\\ \\ Folosim\ formula:\\ \\ \dfrac{2}{n(n+2)}=\dfrac{1}{n}-\dfrac{1}{n+2}[/tex]

[tex]\it a=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\ ...\ +\dfrac{1}{49}-\dfrac{1}{51}=1-\dfrac{1}{51}=\dfrac{50}{51}\\ \\ \\ \dfrac{50}{100}<\dfrac{50}{51}<\dfrac{51}{51} \Rightarrow \dfrac{1}{2}<\dfrac{50}{51}<1 \Rightarrow a\in\Big(\dfrac{1}{2},\ 1\Big)[/tex]

[tex]\it 4.\\ \\ x,\ y>0\ \ \ \ \ (1)\\ \\ 4x+7y=28 \Rightarrow 7x-3x+7y=28 \Rightarrow 7x+7y-28>3x\ \stackrel{(1)}{\Longrightarrow}\ 7x+7y-28>0 \Rightarrow \\ \\ \Rightarrow 7x+7y>28|_{:7} \Rightarrow x+y>4 \Rightarrow 4<x+y\ \ \ \ \ (2)[/tex]

[tex]\it 4x+7y=28 \Rightarrow 4x+4y+3y=28 \Rightarrow 3y=28-4x-4y\ \stackrel{(1)}{\Longrightarrow}28-4x-4y>0 \Rightarrow \\ \\ \Rightarrow 28>4x+4y|_{:4} \Rightarrow 7>x+y \Rightarrow x+y<7\ \ \ \ \ (3)\\ \\ (2),\ (3) \Rightarrow 4<x+y<7\Rightarrow x+y\in(4,\ \ 7)[/tex]