
Zhananmuk
6 год назад
3 ≤ |2х-3| < 7
среднее арифметическое целых решений неравенства
ОТВЕТЫ

Иннокентьевич
Jul 1, 2019
|2x-3|≥3⇒
[2x-3≤-3⇒2x≤0⇒x≤0
[2x-3≥3⇒2x≥6⇒x≥3
-7lt;2x-3lt;7
-4lt;2xlt;10
-2lt;xlt;5
x∈(-2;0] U [3;5)
среднее арифметическое (-1+0+3+4):4=6:4=1,5
[2x-3≤-3⇒2x≤0⇒x≤0
[2x-3≥3⇒2x≥6⇒x≥3
-7lt;2x-3lt;7
-4lt;2xlt;10
-2lt;xlt;5
x∈(-2;0] U [3;5)
среднее арифметическое (-1+0+3+4):4=6:4=1,5
3 ≤ |2х-3| lt; 7
1. 3 ≤ |2х-3|
3 ≤ 2х-3
2xgt;=6
xgt;=3
2x-3lt;=-3
xlt;=0
x=(-00 0] [3 +00))
2. |2х-3| lt; 7
2x-3lt;7
xlt;5
2x-3gt;-7
2xgt;-4
xgt;-2
-------------- (-2) /////////////////////// (5)------------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\[0]--------[3]\\\\\\\\\\\\\\\\\\\\\\\\\\\\
(-2 0] U [3 5)
итаго -1 0 3 4
(-1+0+3+4)/4= 6/4= 3/2=1.5
среднее арифм = 1.5 =3/2
1. 3 ≤ |2х-3|
3 ≤ 2х-3
2xgt;=6
xgt;=3
2x-3lt;=-3
xlt;=0
x=(-00 0] [3 +00))
2. |2х-3| lt; 7
2x-3lt;7
xlt;5
2x-3gt;-7
2xgt;-4
xgt;-2
-------------- (-2) /////////////////////// (5)------------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\[0]--------[3]\\\\\\\\\\\\\\\\\\\\\\\\\\\\
(-2 0] U [3 5)
итаго -1 0 3 4
(-1+0+3+4)/4= 6/4= 3/2=1.5
среднее арифм = 1.5 =3/2
192
Смежные вопросы: