Покажите на координатной плоскости множество решений неравенства: y-x²+6x<0. какая из точек А(3;5) и В(-3;-2) принадлежит множеству решений этого неравенства?

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Ответ:

Qwiq

31.08.2022 19:43

перенесем оба числа в левую стронуи приравняем 0, т.к. это нам даст определение области значений между этими числами

35х^4-6х^8=0

выделим х^4

х^4(35-6х^4)=0

выражение равно0 только когда хотя бы один из множителей равен 0

либо х^4=0

х=0

либо 35-6х^4=0

35=6х^4

х^4=35/6

$x = | \sqrt[4]{ \frac{35}{6} } |$

расставим знаки +/- на графике. если значение принимает положительное зачение, то 35х^4>6х^8, иначе наоборот

ответ

35х^4<6х^8 при

$x < - \sqrt[4]{ \frac{35}{6} }$

35х^4=6х^8 при

$x = - \sqrt[4]{ \frac{35}{6} }$

35х^4>6х^8 при

$x - \sqrt[4]{ \frac{35}{6} }$

но меньше 0

35х^4=6х^8 при х=0

35х^4>6х^8 при

х>0 но меньше

$\sqrt[4]{ \frac{35}{6} }$

35х^4=6х^8 при

$x = \sqrt[4]{ \frac{35}{6} }$

35х^4<6х^8 при

$x \sqrt[4]{ \frac{35}{6} }$

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Ответ:

Эльмаз07

26.04.2021 22:43

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