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Ответ

Проверено экспертом

Ответ дан
7x8

5<a<9

25<5a<45

2<b<7

-7<-b<-2

-7/3<-b/3<-2/3

25-frac{7}{3}<5a-frac{b}{3}<45-frac{2}{3}

25-2frac{1}{3}<5a-frac{b}{3}<44frac{1}{3}

22frac{2}{3}<5a-frac{b}{3}<44frac{1}{3}

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a^2+10b^2geq 6ab\\a^2-6ab+10b^2geq 0\\a^2-6ab+9b^2+b^2geq 0\\(a-3b)^2+b^2geq 0

неравенство верно при любом значении a и b

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