Question

Problema din imagine.​

Answer 1

Răspuns:

Explicație pas cu pas:

(n-1)(n-2)=n²-3n+2=n²+2n+1-2n-3n-1+2=(n+1)²-5n+1=(n+1)²-5n+5-5+1=

=(n²+1)²-5(n+1)+6

[(n²+1)²-5(n+1)+6]/(n+1)=(n+1)-5+6/(n+1)

D6 naturali=1,2,3,6

n+1=6      n=5

n+1=3       n=2

n+1=2       n=1  

n+1=1        n=0    ⇒S={0,1,2,5}      

Verfificam  n=0     (-1)(-2)/1=2

                  n=1        E(n)=0

                   n=2      E(2)=0

                   n=5     E(n)=4·3.4=3

Answer 2

$\it (n-1)(n-2)=n^2-2n-n+2=n^2-3n+2\\ \\ \dfrac{(n-1)(n-2)}{n+1}=\dfrac{n^2-3n+2}{n+1}\in\mathbb{N}\ \Rightarrow\ n+1|n^2-3n+2\ \ \ \ \ \ (1)\\ \\ \\ Dar,\ \ n+1|n+1 \Rightarrow\ n+1|(n+1)\cdot n \Rightarrow\ n+1|n^2+n\ \ \ \ \ \ \ \ \ (2)$

$\it (1),\ (2) \Rightarrow\ n+1|n^2+n-n^2+3n-2 \Rightarrow\ n+1|4n-2\ \ \ \ \ (3)\\ \\ Dar, n+1|n+1 \ \Rightarrow\ n+1|(n+1)\cdot4\ \Rightarrow\ n+1|4n+4\ \ \ \ \ (4)\\ \\ (3),\ (4) \Rightarrow\ n+1|4n+4-4n+2 \Rightarrow\ n+1|6 \Rightarrow\ n+1\in\{1,\ 2,\ 3,\ 6\}_{-1} \Rightarrow\ \\ \\ \Rightarrow\ n\in\{0,\ \ 1,\ \ 2,\ \ 5\}$