Question

3a/2=4b/3=5c/4
15a-4b+5c=4
Aratati ca are loc inegalitatea 4/3<3a-3b+5c<7/5

Answer 1

primele 2 relatii alcatuiesc de fapt un sistem de 3 ecuatii(avem atate ecuatiicate semnede egal avem)  cu 3 nec.(a; b;c)  de grad 1
deci il rezolvam
 folosim metoda substitutiei
b=9a/8
c=12a/10=6a/5

15a-4*(9a/2)+5*(6a/5)=4
15a-18a+6a=4
3a=4
a=4/3
b=(9/8)*(4/3)=3/2
c=(6/5)*(4/3)=8/5

3a-3b+5c=38(4/3)-3*(3/2)+5*(8/5)=4-9/2+8=-0,5+8=7,5

4/3<7,5, adevarat
dar 7,5<7/5 e fals
 probabil relatia este≤7,5
TEXT PROBLEMA GRESIT
Demonstratie;
  asa cum e scris  texul ,  ar trebui sa avem
4/3<d<7/5  ar insemna sa avem
1.(3)<d<1,2
1,(3)<1,2 FALS,∀d∈R