Question

a Cate cifre are numarul
X=12345678910111213141516171819202122232425
b Cate cifre are numarul
X=12345678910111213141516171819....198199200
c Determinati x care verifica egalitatile
x+89=115 x=?
2x+23=91 x=?
327-3x=15+48 x=?
15+(x+2)+21=47
2x+37=79-14
32+(2x-5)=63+14

Answer 1

a)
1 ---> 9 sunt 9 cifre
10 ---> 25 ; (25-10+1)×2 = 16×2 = 32 cifre
9 + 32 = 41 cifre (are numarul)

b)
1--->9 sunt 9 cifre
10 ---> 99 ; (99-10+1)×2 = 90×2 = 180 cifre
100 ---> 200 ; (200-100+1 )×3 = 101×3=303 cifre
9+180+303 = 492 cifre (are numarul)

c)
X+89 = 115
X = 115-89
X = 26

2X+23=91
2X = 91-23
2X = 68
X = 68:2
X = 34

327-3x=15+48
327-3X = 63
3X = 327-63
3X = 264
X = 264 : 3
X = 88

15+(x+2)+21=47
X + 38 = 47
X = 47-38
X = 9

2x+37=79-14
2X = 65-37
2X = 28
X = 28:2
X = 14

32+(2x-5)=63+14
2X-5 = 77-32
2X-5 = 45
2X = 45+5
2X = 50
X=50:2
X=25