Vă rog frumos ,știe cineva cum se face? E urgent ,la ora 8 este termenul limită!
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Răspuns:
A={x∈N∣2x+163∈N}
\frac{63}{2x + 1} \:\in\:\mathbb{N} = > 2x + 1 \: \in \: D_{63_{+}}2x+163∈N=>2x+1∈D63+
D_{63_{+}}=\left\{1,3,7,9,21,63\right\}D63+={1,3,7,9,21,63}
1)2x + 1 = 11)2x+1=1
2x = 1 - 12x=1−1
2x = 02x=0
x = 0 \: \in \: \mathbb{N}x=0∈N
2)2x + 1 = 32)2x+1=3
2x = 3 - 12x=3−1
2x = 2 \: | \div 22x=2∣÷2
x = 1 \: \in \: \mathbb{N}x=1∈N
3)2x + 1 = 73)2x+1=7
2x = 7 - 12x=7−1
2x = 6 \: | \div 22x=6∣÷2
x = 3 \: \in\: \mathbb{N}x=3∈N
4)2x + 1 = 94)2x+1=9
2x = 9 - 12x=9−1
2x = 8 \: | \div 22x=8∣÷2
x= 4 \: \in \: \mathbb{N}x=4∈N
5)2x + 1 = 215)2x+1=21
2x = 21 - 12x=21−1
2x = 20 \: | \div 22x=20∣÷2
x = 10 \: \in \: \mathbb{N}x=10∈N
6)2x + 1 = 636)2x+1=63
2x = 63 - 12x=63−1
2x = 62 \: | \div 22x=62∣÷2
x = 31 \: \in \: \mathbb{N}x=31∈N
= > x \: \in \: \left\{0,1,3,4,10,31\right\}=>x∈{0,1,3,4,10,31}
= > A=\left\{0,1,3,4,10,31\right\}=>A={0,1,3,4,10,31}
Sper că te-am ajutat