1) log(0,2)(4-2x)>log(0,2)(0,2⁻¹)
ОДЗ: 4-2x>0
4>2x
x<2
4-2x<5
-2x<1
x>-0,5
x∈(-0,5;2)
2) ОДЗ: x>0
5>x
x∈(0;5)
x>5-x
2x>5
x>2,5
x∈(2,5;5)
3) ОДЗ:
(x+1)/(2x-1)>0
x∈(-∞;-1)∪(0,5;+∞)
log(0,25)(x+1/2x-1)≥log(0,25)(0,25⁻¹)
(x+1)/(2x-1)≤4
(x+1-8x+4)/(2x-1)≤0
(5-7x)/(2x-1)≤0
x∈(-∞;0,5)∪[5/7;+∞)
x∈(-∞;-1)∪[5/7;+∞)
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Verified answer
1) log(0,2)(4-2x)>log(0,2)(0,2⁻¹)
ОДЗ: 4-2x>0
4>2x
x<2
4-2x<5
-2x<1
x>-0,5
x∈(-0,5;2)
2) ОДЗ: x>0
5>x
x∈(0;5)
x>5-x
2x>5
x>2,5
x∈(2,5;5)
3) ОДЗ:
(x+1)/(2x-1)>0
x∈(-∞;-1)∪(0,5;+∞)
log(0,25)(x+1/2x-1)≥log(0,25)(0,25⁻¹)
(x+1)/(2x-1)≤4
(x+1-8x+4)/(2x-1)≤0
(5-7x)/(2x-1)≤0
x∈(-∞;0,5)∪[5/7;+∞)
x∈(-∞;-1)∪[5/7;+∞)