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2014-06-05T18:07:12+03:00
Echiunitara⇒x+3=5, x=2
subunitara x+3<5, x<2
 \frac{x+3}{5} = \frac{21}{15}
15(x+3)=21*5
15x+45=105
15x=60
x=4
 \frac{x+3}{5} <3/2
2(x+3)<15
2x+6<15
2x<9
x<9/2
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2014-06-05T18:09:21+03:00
A)(x+3)/5= 5/5

x+3=5
x=5-3
x=2

b)(x+3)/5=4/5
x∈N
x+3=4
x=4-3
x=1

c)(x+3)/5=21/15  aducem la acelasi numitor
3(x+3)/15= 21/15
3(x+3)=21
3x+9=21
3x=21-9
3x=12
x=12:3
x=4

d) (x+3)/5 < 3/2  aducem la acelasi numitor
2(x+3)/10 < 15/10

2(x+3) < 15
2x+6 < 15
2x < 15-6
2x < 9
x < 9/2 
x < 4,5  ⇒x∈{0,1,2,3,4}
x∈N

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