Question

aflati patru numere intregi consecutive al caror produs este 5040

Answer 1

[tex]x(x+1)(x+2)(x+3)=5040\\ (x^2+3x)(x^2+x+2x+2)=5040\\ (x^2+3x)(x^2+3x+2)=5040\\ Notam:x^2+3x=a\\ a(a+2)=5040\\ a^2+2a=5040|+1\\ a^2+2a+1=5041\\ (a+1)^2=5041\\ |a+1|=71\\ De\ aici\ avem\ doua\ posibilitati:\\ a)a+1=71\\ x^2+3x=70\\ x^2+3x-70=0\\ x^2+10x-7x-70=0\\ x(x+10)-7(x+10)=0\\ (x+10)(x-7)=0=\ \textgreater \ x_1=-10,x_2=7 b)a+1=-71\\ x^2+3x=-72\\ x^2+3x+72=0\\ \Delta=9-4*72\\ \Delta=9-288=-279=\ \textgreater \ x\ nu\ apartine\ lui\ R.\\ Deci\ solutia\ va\ fi:\\ S:(a,b,c,d)\ ap.\ \{(-10;-9;-8;-7),(7;8;9;10)\}[/tex]