Question

Exercițiul 3 sau 4
Dau coroana!

Answer 1

3.  41·n! +n! /(n-(n-1))! =(n+2)! <=> 41·n! +n!=(n+2)! <=> 42·n!=(n+2)! <=> 42·n!=n!·(n+1)·(n+2) dar deoarece pentru orice n∈N => n!≠0 <=> (n+1)·(n+2)=42 <=> n+1 < n+2 iar 42=6·7 =>n+1=6 si n+2=7 <=> n=5 .

4.  f(x)=ax² +bx +c ,unde a,b,c∈R si a≠0 a.i. x∈R ,iar in cazul nostru avem

f(0)=-3 <=> 0+0+c=-3 <=> c=-3 ;

f(1)=-6 <=> a+b+c=-6 <=> a+b=-3;

f(2)=-5 <=> 4a+2b+c=-5 <=> 4a+2b=-2 <=> 2a+b=-1 .

2a+b=a+(a+b)=-1 <=> a=2 => b=-5 de unde obtinem functia de gradul al II-lea

f(x)=2x² -5x -3 .