Răspuns:
1. x/5=y/7=k => x=5k si y=7k
(x+y)/2=29,4 => x+y=58,8
Inlocuim si avem 5k+7k=58,8 => 12k=58,8 => k=4,9 deci x=24,5 si y=34,3
2. x/2=y/3=k => x=2k si y=3k
dar [x,y]=36 => [2k,3k]=36 => k=6 deci x=12 si y=18
3.semiperimetrul = suma laturilor impartita la 2 => suma laturilor (a+b+c = 126 )
0,(3)=3/9=1/3
0,5=1/2
0,2=1/5
laturile sunt invers proportionale cu 1/3 , 1/2, 1/5 => a/3 = b/2 = c/5 = k =>
=>a=3k
b=2k
c=5k
3k+2K+5k=126=>10k=126=>k=12,6
a=3*12,6=37,8
b=2*12,6=25,2
c=5*12,6=73
4.
x/0,(3) = y/0,1(6) => 3x = 8y => x/6 = y/3;
y*15 = z*9 => y/3 = z/5;
atunci x/8 = y/3= z/5 = k => x = 6k; y = 3k; z = 5k; unde k e nr. natural nenul;
cea mai mica suma S se obtine pentru k = 1 => x = 6, y = 3 si z = 5 => S = 14;
testul 2
x+y+z =20
x/0.2 =y/0.6 =z/1.2
x/0.2+y/0.6+z/1.2 =x+y+z/0.2+0.6+1.2 =20/2
=10
x/0.2 =10 ;
y/0.6 =10 ;
z/1.2 =10 ;
х =10*0.2 =2
y =0.6*10 = 6
z =1.2*10 =12
2.mii lene
3.
0,(3)= 3/9=1/3
0,25= 25/100=1/4 0,2= 2/10= 1/5
(a+b); (b+c); (c+a) ip cu 1/3 ; 1/4; 1/5
(a+b) * 1/3 = (b+c) * 1/4 (c+a) * 1/5 =k
(a+b)/3= (b+c)/4 = (c+a)/5=k
a+b=3k
b+c=4k
c+a=5k
2a+2b+2c=3k+4k+5k
2(a+b+c)= 12k 1:2
a+b+c=6k
C= 6k-3k C=3k
a=6k-4k
a=2k
b=6k-5k b=k
a3 * 2b4 * c? =72
(2k)? * 2k* * (3k)? =72
4k? * 2k* * 9k? =72
72k =72
k=1 =>k=1
a=2*1=2
b=1
c=3*1=3
4.
5a = 6b = 10c b = 5a/6 (а? + b? +с? )y (abc) = 7/9 a? + b? + с? =а?+ 25а? /36 + а? 14%3 (36а? + 25a* + 9а*)/36 = 70a/36 = 35a°/18 abc = a.5а/6 al2 = 5a3 /12 2/a = 1/3
35а? / 18 - 12/ 5a3 = 7/9 7:21 За = 7/9
a = 6 b= 5 с = 3
