Доказать что функция f(x)=-4x^3-2x+5 убывает на всей числовой оси

=8

=87a + 5b = 8a

=87a + 5b = 8a5b = 8a - 7a

=87a + 5b = 8a5b = 8a - 7aa = 5b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b =

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b =

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b =

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b

=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b =32

Перевод:  классифицируйте эти числа на простые множители.

На математическом языке разложите.

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 713  I 23

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 217  I 7

    31 I 31

3751 I 11                3751=11*11*31

 341 I 11

    31 I 31

4805 I 5               4805=5*31*31

 961  I 31

    31  I 31

1225  I  5               1225=5*5*7*7

 245 I 5

   49 I 7

      7 I 7

961 I 31                  961=31*31

  31 I 31

11 495 I 5               11495=5*11*11*19

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19 855 I 5                19855=5*11*19*19

  3971  I 11

    361  I 19

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47 096 I 7               47 096=2*2*2*7*29*29

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1914 I 2                      1914=2*3*11*29

957 I 3

319  I 11

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133  I 7

 19  I 19

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551  I 19

   29 I 29

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