Question

E2 a, b, c și d. Va rog sa mă ajutați

Answer 1

Răspuns:

Explicație pas cu pas:

E2 a, b, c și d.

a) (A₇⁴-P₅)/A₅²=[7!/(7-4)!-5!]/5!/(5-2)!=(7!/3!-5!)/(5!/3!)=[(1·2·3·4·5·6·7/1·2·3)-1·2·3·4·5]/(1·2·3·4·5/·2·3)=(4·5·6·7-1·2·3·4·5)/(4·5)=

(4·5)(6·7-2·3)/(4·5)=67·3=42-6=36

b) (2P₃+3A₄²)/(5P₅-P₃)=[(2·3!+3·4!/(4-2)!]/(5·5!-3!)=(2·1·2·3+3·1·2·3·4/1·2)/(5·1·2·3·4·5-1·2·3)=(2·1·2·3+3·3·4)/(5·1·2·3·4·5-1·2·3)=

3·4(1+3)/2·3(5·4·5-1)=2·4/99=8/99    

c) (A₁₁⁶-A₁₁⁵)/(A₁₀⁵-A₁₀⁴)=(11!/5!-11!/6!)/(10!/5!-10!/6!)=

(1·2·3·4·5·6·7·8·9·10·11/1·2·3·4·5-1·2·3·4·5·6·7·8·9·10·11/1·2·3·4·5·6)/(10!/5!-10!/6!)=

(6·7·8·9·10·11-7·8·9·10·11)/(1·2·3·4·5·6·7·8·9·10/1·2·3·4·5-1·2·3·4·5·6·7·8·9·10/1·2·3·4·5·6)=

7·8·9·10·11(6-1)/(6·7·8·9·10-7·8·9·10)=5·7·8·9·10·11/7·8·9·10(6-1)=5·11/5=11

d) (A₉³+A₈⁴)/A₇³=[9!/(9-3)!+8!/(8-4)!]/7!/(7-3)!=(9!/6!+8!/4!)/(7!/4!)=

(1·2·3·4·5·6·7·8·9/1·2·3·4·5·6+1·2·3·4·5·6·7·8/1·2·3·4)/1·2·3·4·5·6·7/1·2·3·4=

(7·8·9+5·6·7·8)/5·6·7=7·8(9+5·6)/5·6·7=8·39/5·6=4·39/5·3=4·13/5=52/5