ОДЗ: х - 3 > 0 и х - 1 > 0, т.е. х > 3 и х > 1, значит, х > 3
log₃(x - 3) ≤ log₃3 - <span>log₃(x - 1)
</span>log₃(x - 3) + log₃(x - 1) ≤ <span>log₃3
</span>log₃((x - 3)(x - 1)) ≤ <span>log₃3
</span>(x - 3)(x - 1) ≤ 3
x² - x - 3x + 3 - 3 ≤ 0
x² - 4x ≤ 0
x(x - 4) ≤ 0
+ - +
______|____________|___________
0 4
С учетом ОДЗ: х > 3 и 0 ≤ х ≤ 4 получим ответ: х ∈ (3; 4].
|-15|=-(-15)=15
|15|=15
2х-7=-15
2х=-8
х=-4
2х-7=15
2х=23
х=11.5
Sin2x+2sinx=cosx+1
2sinx*cosx+2sinx-cosx-1=0
(2sinx-1)(cosx+1)=0
1.cosx=-1
x=π+2πn
2. sinx=1/2
x=(-1)^n*π/6+πn
(3а – 5) + (2а – 4) – (5а - 2)=3а – 5 + 2а – 4 – 5а + 2=5а – 5 – 4 – 5а + 2=<span>– 5 – 4 + 2=-7
</span>4(3а - 3) – 5 (4а + 4) + 8а =12а - 12 – 20а -20 + 8а =20а - 12 – 20а -20 = - 12 -20<span> =-32</span>
1. an = a1+d(n-1)
a11=-2,4 +0,8(11-1)
a11=-2,4+8,8-0,8
a11=5,6