Question

Va rog repede. Dau 60 de puncte


Media aritmetica a numerelor a,b,c este 4+8(radical)2.

a) Stiind ca a=3+radical32, b=5+radical72, aflati numarul

b) Determinati cele trei numere, daca 2a=3b=6c.

Answer 1

Răspuns:

Explicație pas cu pas:

(a+b+c)/3= 4+8 radical 2

a+b+c= 3 ori (4+8 radical 2)

a+b+c=12 + 24 radical 2

a) a= 3 + radical 32

Radical 32=4 radical 2

Din astea doua=> a= 3 + 4 radical2

b= 5+ radical 72

Radical 72= 6 radical 2

Dibastea doua => b= 5+6 radical 2

Acum avem:

a+b+c= 12+24 radical 2

Si: a= 3+4 radical 2

b= 5+6 radical 2

Inlocuim mai sus si vom avea

a+b+c=12+24 radical 2

3+ 4 radical 2+ 5+6radical 2+c= 12+24 radical 2

8+ 10 radical 2+c= 12+ 24 radical 2

C= 12+24 radical 2 - 8-10 radical 2

C= 4 +14 radical 2

b) a+b+c= 12+24 radical 2

2a=3b=6c

2a=3b=> a= 3b/2

Inlocium mai sus si vom avea

3b/2+ b+c= 12+ 24 radical 2

(3b+2b)/2+c= 12+24 radical 2

5b/2+c= 12+24 radical 2

3b=6c=> c= 3b/6

Inlocium mai sus si vom avea

5b/2+ 3b/6= 12+ 24 radical 2

(15b+3b)/6= 12+24 radical 2

18b/6= 12+ 24 radical 2

3b= 12+ 24 radical 2

b= (12+ 24 radical 2)/3

b= 3 ori (4+ 8 radical 2)/3

b= 4+ 8 radical 2

c= 3b/6=> c= [3 ori (4+ 8radical 2)/6]

c= (4+ 8 radical 2)/2 = 2 ori (2+ 4 radical 2)/2= 2+ 4 radical 2

a= 3b/2=> a= [3 ori (4+ 8 radical 2)/2]

a= (12+ 24 radical 2)/2= 2 ori ( 6+ 12 radical 2)/2= 6+12 radical 2

Pentru punctul b) Mai este o metoda de rezolvare cu k. Iata metoda

2a+3b+6c=k

2a=k=> a=k/2

3b= k=> b= k/3

6c= k=> c= k/6

a+ b+c=12+ 24 radical 2

K/2+ k/3+k/6= 12+24 radical 2

(3k+ 2k+k)/6= 12+ 24 radical 2

6k/6= 12+24 radical 2

K= 12+24 radical 2

a= k/2=> a= (12+ 24 radical 2)/2= 2 ori (6+ 12radical 2)/2= 6+ 12 radical 2

b= k/3=> b= 12+24 radical 2/3= 3 ori (4+ 8 radical 2)/3= 4+ 8 radical 2

c= k/6=> c=( 12+24 radical 2)/6= 6 ori (2+ 4 radical 2)/6= 2+ 4 radical 2