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Vladka15
@Vladka15
June 2022
1
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составьте уравнение касательной к графику функции у=f(х) в точке с абсциссой х=а
f(х)=2х-5/5-х, а=4
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malovik11
Uhjibfhhgvuovdrtgvjidtttvovxtybvofx
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Vironuka
F'(x)=10*2х+2х-5/(5-х)2=5/(5-х)2
Vironuka
у=3+5(4-х)= 5х-17 - вот уравнение касательной
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