Question

Ajutor!Ex 38 si 39.Din poza!Pt a 6a.

Answer 1

38)
a+b+c=84
a/1,(4)=b/1,(5)=c/1,(6)=r
r este o variabila aleasa la intamplare, ca sa scoatem a, b si c in functie de ea.

1,(4)= 13/9
1,(5)= 14/9
1,(6)= 15/9

Deci egalitatea aia devine: 9a/13=9b/14=9c/15=r deci:
a=13/9*r
b=14/9*r
c=15/9*r
Rezulta ca 42/9*r=84. De aici r=9*84/42, deci r=18. Si de aici calculezi a,b,c.
a=26 b=28 c=30

Dupa aceeasi metoda faci si 39. Introduci o variabila auxiliara dupa care sa le scoti pe celelalte
39. x+2y+3z=78 x/3=y/4=z/5=r => x=3r y=4r z=5r Deci: 3r+8r+15r=78 => 26r=78 => r=78/26 => r=3 Deci x=3*3=9 y=4*3=12 z=5*3=15

Answer 2

   
[tex]\displaystyle\\ 38)\\ a,~b,~c~~\text{ d.p. cu }~~1,\!(4),~1,\!(5),~\!(6)\\ a+b+c=84\\ \text{Rezolvare:}\\\\ \frac{a}{1,\!(4)} = \frac{b}{1,\!(5)} = \frac{c}{1,\!(6)} =k\\\\ a=1,\!(4)k;~~b=1,\!(5)k;~~c=1,\!(6)k\\\\ a+b+c=184\\\\ 1,\!(4)k+1,\!(5)k+1,\!(6)k=84\\\\ k(1,\!(4)+1,\!(5)+1,\!(6))=84\\\\ k\left(\frac{14-1}{9}+\frac{15-1}{9}+\frac{16-1}{9}\right)=84\\\\ k\left(\frac{13}{9}+\frac{14}{9}+\frac{15}{9}\right)=84\\\\ k\cdot\frac{13+14+15}{9}=84\\\\ k\cdot\frac{42}{9}=84\\ k=84\cdot\frac{9}{42}=\boxed{18}[/tex]

[tex]\displaystyle\\ a=1,\!(4)k= \frac{13}{9}k= \frac{13}{9}\cdot 18= 13\cdot 2=\boxed{\bf 26}\\\\ b=1,\!(5)k= \frac{14}{9}k= \frac{14}{9}\cdot 18= 14\cdot 2=\boxed{\bf 28}\\\\ c=1,\!(6)k= \frac{15}{9}k= \frac{15}{9}\cdot 18= 15\cdot 2=\boxed{\bf 30}\\\\ \text{Verificare:}\\ 26+28+30 = 84[/tex]


[tex]\displaystyle\\ 39)\\ x,~y,~z~~~\text{ d.p. cu }~~~3,~4,~5\\ x+2y+3z=78\\\\ \text{Rezolvare:}\\\\ \frac{x}{3} =\frac{y}{4} =\frac{z}{5} =k\\\\ x=3k\\ y=4k\\ z=5k\\\\ x+2y+3z=78\\\\ 3k+2\cdot 4k+3\cdot 5k=78\\\\ 3k+8k+15k=78\\\\ 26k=78\\\\ k = \frac{78}{26} =3\\\\ x=3k=3\cdot 3=\boxed{9}\\ y=4k=4\cdot 3=\boxed{12}\\ z=5k=5\cdot 3=\boxed{15}\\ \text{Verificare:}\\ x+2y+3z=9+2\cdot12+3\cdot 15=9+24+45=78[/tex]