Question

Rog urgent folosind exemple propuse in poze sa ma ajutati la rezolvarea exercițiilor (1342, 1344, 1346,1348,1351)

Answer 1

Răspuns:

1342

$log_{\frac{1}{2} } (2x+3)>0$

(2x+3)>0

x>$-\frac{3}{2}$

x∈(-$\frac{3}{2}$,+∞)=DVA

(2x+3)>0

x>$-\frac{3}{2}$

x∈($-\frac{3}{2} ,$+∞)

2x+3<1

2x<-2

x<-1

x∈($-\frac{3}{2} ,-1)$⊂DVA

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1344

log₃(1-2x)≥log₃(5-2x)

1-2x>0

-2x>-1

x<$\frac{1}{2}$

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: