Найдите значения функции: f(x) = x² + 1/x в точках -2; 5; 10. Created by korotkovmaks04. algebra-ru

x в точках -2; 5; 10...

$f(x)=\frac{x^{2}+1 }{x}\\\\f(-2)=\frac{(-2)^{2}+1 }{-2}=-\frac{4+1}{2}=-\frac{5}{2}= -2,5\\\\f(5)=\frac{5^{2}+1 }{5}=\frac{25+1}{5}=\frac{26}{5}=5,2\\\\f(10)=\frac{10^{2}+1 }{10}=\frac{100+1}{10}=\frac{101}{10}=10,1$

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Ответ:

f(-2)=(-2)²+1/(-2)=3,5

f(5)=5²+1/5=25,2

f(10)=10²+1/10=100,1

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