-(x+3)²+2(4-x)*(x+3)=(x+3)(-x-3+8-2x)=(x+3)(5--3x)
htibv ythfdtycndj
(x+3)(5-3x)>0
отметим на координатном луче точки х=-3 и х= 5/3=
1 2/3
х∈(-3; 5/3)
0.7y - 0.2x + 0.3y +0.2x = 1y = -0.14
6) 5(1 + 2x) - (2x - 1) = 7
5 + 10x - 2x + 1 = 7
10x - 2x = 7 - 5 - 1
8x = 1
x = 1 : 8 = 0,125
7) 11(1 - 2x) - 8(3x - 5) = 1 + 4x
11 - 22x - 24x + 40 = 1 + 4x
- 22x - 24x - 4x = 1 - 11 - 40
- 50x = - 50
x = - 50 : (- 50) = 1
Решение
log₂² x - 4log₂ x + 3 = 0
ОДЗ: x > 0
log₂ x = t
t² - 4t + 3 = 0
t₁ = 1
t₂ = 3
1) log₂ x = 1
x = 2¹
x₁ = 2
2) log₂ x = 3
x = 2³
x₂ = 8
Решение смотрите во вложении....