Решите пожалуйста) a) p(p+7)>7p-1 б) 8y(3y-10)

August 2022 1 5 Report

Решите пожалуйста)
a) p(p+7)>7p-1
б) 8y(3y-10)<(5y-8)^2
в) (a-3)(a-5)<(a-4)^2
г) f(x) = -5x+6
f(x)= -3
f(0)= 0

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Apofeoz

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$a)\ p(p+7)\ \textgreater \ 7p-1,\\p^2+7p\ \textgreater \ 7p-1,\\p^2+7p-7p+1\ \textgreater \ 0,\\p^2+1\ \textgreater \ 0,\\p^2\ \textgreater \ -1,\\p\in \mathh {R};\\\\b)\ 8y(3y-10)\ \textless \ (5y-8)^2,\\24y^2-80y\ \textless \ 25y^2-80y+64,\\-y^2\ \textless \ 64,\\y^2\ \textgreater \ -64,\\y\in \mathh {R};\\\\c)\ (a-3)(a-5)\ \textless \ (a-4)^2,\\a^2-5a-3a+15\ \textless \ a^2-8a+16,\\15\ \textless \ 16,\\a\in \mathh {R};\\\\d)\ f(x)=-5x+6,\\f(x)=-3,\\-5x+6=-3,\\-5x=-9,\\x=\frac{9}{5}=1\frac{4}{5}=1.8,\\f(1.8)=-3;\\f(0)=-5\bullet0+6=6,\\f(0)=6.$

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