Question

pls cele încercuit pun 100 puncte

Answer 1

28)
(a+b+c):3=16
(a+b+c)=16x3⇔a+b+c=48 (suma celor 3 nr)
(a;b;c) d.p. (4;5;7)
a/4=b/5=c/7=K ⇔a=4K
                            b=5K
                            c=7K
4K+5K+7K=48
16K=48
K=48:16
K=3
a=3x4⇔a=12
b=3x5 ⇔b=15
c=3x7 ⇔c=21

30)
2,(4)=2 intregi 4/9=  22/9
a,b  ip cu 3  si 22/9
3a=22b/9=k
a=k/3
22b=9k
b=9k/22
ma=(a+b)/2=263
a+b=263*2
a+b=526
k/3+9k/22 =526  aducem la acelasi numitor
22k+27k =34716
49k=34716
k= 34716/49
a=34716/49 /2=  34716/49  *1/2=  17358/49
b= 9*34716/49/22=  9*34716/49 *1/22= 9*1578/49=  14202/49

31)
Fie a, b, c, d partile  invers proportionale cu nr 2,3, 0,5 si 0, (3)
adica a*2=b*3=c*0,5=d*3/9=k
a=k/2
b=k/3, 
c=k/0,5=10k/5=2k
d=9k/3=3k
si a+b+c+d=7
inlocuind am
k/2+k/3+2k+3k=7
numitorul comun e 6
3k+2k+30k=42
35k=42
5k=6 ⇒ k=6/5
a= k/2=6/5*1/2=6/10=3/5
b=k/3=6/15=2/5
c=2k=6*2/5=12/5
d=3k=6*3/5=18/5