Question

Va rog urgent ex : 24,26,28,29

Answer 1

24.
a)Numerele prime luate in discutie sunt 11, 13, 17, 19.
A={22, 26, 34, 38}
b)Numerele care indeplinesc conditia divizibilitatii sunt 114, 144, 174.
A={119/114, 119/144, 119/174}

26.
2/3 = 4/6 = 10/15
Ca sa dea ultima echivalenta, trebuie sa amplifici cu 2,5...
=>x+1=4 =>x=3
=>2y+1=15 =>2y=14 =>y=7

27.
20/30=2/3
250/350=5/7
(2x3x7)/(3x7x11)=2/11
2³/2⁵=1/2²=1/4
(2⁷x3²x5⁴)/(2⁷x3³x5²x11)=5²/3x11=25/33
5200/18200=26/91
1716/4290=2/5
72a/108a=2/3
34(a+b)/51(a+b)=2/3

28.
a)x=2, y=4
b)x=9, y=5
c)(7x+21):y=x+3 =>y=7(x+3)/(x+3) =>y=7
  (x+2):y=5 =>y=(x+2)/5 =>x+2=35 =>x=33

29.
24/36 = 12/18 = 4/6 =>x+2=18 =>x=16
                                =>y+1=4 =>y=3

Answer 2

24. a. Numerele prime de forma 1a in baza 10 sunt {11;13;17;19}
1a /x=1/2 ⇔x=2 ·1a ⇒x∈{22;26;34;38}
A∈{22;26;34;38}
b. Un numar natural nenul este divizibil cu 6 daca
-are ultima cifra 0;2;4;6 sau 8;
-suma cifrelor numarului respectiv formeaza un alt numar multiplu de 3;
Avem 1a4 (in baza 10) divizibil cu 6 daca
1+4+a=5+a-divizibil cu 3
a∈{1;4;7}
Numerele gasite sunt:114;144;174.
x=119/1a4 ⇒x∈{119/114;144/174;119/174}
A∈{119/114;119/174}
26. 2/3=x+1 /6 ⇔
3(x+1)=12 ⇔
x+1=4 ⇔
x=3;
x+1 /6=10/2y+1 ⇔
4/6=10/2y+1 ⇔
60=4(2y+1) ⇔
8y+4=60 ⇔
8y=56 ⇔
y=7
In concluzie x=3; y=7.
28.a. 6/8=3/y ⇔
24=6y ⇔
y=4;
6=3x ⇔
x=2;
b. x+1/15=2/3 ⇔
30=3(x+1) ⇔
x+1=10 ⇔
x=9;
10=2x ⇔
x=5;
c. 7x+21/x+2=x+3/5 ⇔
5(7x+21)=(x+2)(x+3) ⇔
35(x+3)=(x+2)(x+3) ⇔
x+2=35 ⇔
x=33;
252=36y ⇔
y=7;
29. 24/16=12/x+2 ⇔
192=24(x+2) ⇔
x+2=8 ⇔
x=6;
12/8=y+2/6 ⇔
8(y+2)=72 ⇔
y+2=9 ⇔
y=7;