Răspuns:
1342
[tex]log_{\frac{1}{2} } (2x+3)>0[/tex]
(2x+3)>0
x>[tex]-\frac{3}{2}[/tex]
x∈(-[tex]\frac{3}{2}[/tex],+∞)=DVA
(2x+3)>0
x>[tex]-\frac{3}{2}[/tex]
x∈([tex]-\frac{3}{2} ,[/tex]+∞)
2x+3<1
2x<-2
x<-1
x∈([tex]-\frac{3}{2} ,-1)[/tex]⊂DVA
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1344
log₃(1-2x)≥log₃(5-2x)
1-2x>0
-2x>-1
x<[tex]\frac{1}{2}[/tex]
x∈(-∞, 1/2)
5-2x>0
-2x>-5
x<5/2=2,5
x∈(-∞; 2,5)
x=(-∞, 1/2)∩(-∞, 2,5)=(-∞,1/2)=DVA
Delogaritmezi
1-2x≥5x-2
-2x-5x≥-2-1
-7x≥ -3
x≤3/7
x∈(-∞, 3/7]⊂DVa
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1348
log₂(3x-2)>log₂(6-5x)
3x-2>0
x>2/3
x∈(2/3,+∞)
6-5x>0
-5x>-6
x<6/5
x∈(-∞, 6/5)
x∈(2/3,+∞)∩(-∞,6/5)=(2/3,6/5)
Delogaritmezi
3x-2>6-5x
3x+5x>6+2
8x>8
x>1
x∈(1,+∞)⊂DVA
Explicație pas cu pas:
