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Angelicus
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Sa se rezolve sistemul de ecuatii:

[tex] \frac{x!}{(y+1)!(x-y-1)!}= \frac{5x}{2} [/tex]

[tex] \frac{(x-1)!}{y!(x-y-1)!}=10 [/tex]

Mersi

Răspuns :

Hell0
2x!=5x(y+1)!(x-y-1)!
2x(x-1)!=5x(y+1)!(x-y-1)!


2(x-1)!=5y(y+1)(x-y-1)!
  (x-1)!=10y!(x-y-1)!

=> 2*2*5y!(x-y-1)!=5y!(y+1)(x-y-1)!
4=y+1 => y=3

(x-1)!=10y!(x-y-1)!
(x-1)!=10*3!(x-4)!
(x-4)!(x-3)(x-2)(x-1)=60(x-4)!
(x-3)(x-2)(x-1)=60
x^3-6x^2+11x-66=0
(x-6)(x^2+11)=0 => x=6







Maryanabel3579
[tex] \frac{x!}{(y+1)!(x-y-1)!} = \frac{5x}{2} il scriem \frac{(x-1)!x}{y!(y+1)(x-y-1)!} = \frac{5x}{2} [/tex]
si il impartim la: [tex] \frac{(x-1)!}{y!(x-y-1)!} =10[/tex]
si obtinem:
[tex] \frac{(x-1)!x}{y!(y+1)(x-y-1)!} : \frac{(x-1)!}{y!(x-y-1)!} = \frac{5x}{2}:10 [/tex]
[tex] \frac{x}{y+1} = \frac{x}{2} * \frac{1}{2} = \frac{x}{4} [/tex]

vine:
4x=xy+x
4x-x=xy  |:x
3=y => y=3




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