Question

При каких целых значениях n дробь (n^2-n+3)/(n+1) является целым числом

Answer 1

$\frac{n^2-n+3}{n+1}= \frac{n^2+(-2n+n)+(-2+5)}{n+1} = \frac{(n^2-2n)+(n-2)+5}{n+1}=\\\\= \frac{n(n-2)+(n-2)+5}{n+1}= \frac{(n+1)(n-2)+5}{n+1}= \frac{(n+1)(n-2)}{n+1}+ \frac{5}{n+1}=n-2+ \frac{5}{n+1}\\\\n\in Z\; =>\; n-2\in Z$

Далее, должны одновременно выполняться два условия:
1) |n+1|≤5    и      2) 5/(n+1)∈Z
-5≤n+1≤5
-5-1≤n≤5-1
-6≤n≤4
Из данного промежутка подходят лишь 4 варианта:
n=-6     (5\(-6+1)=5\(-5)=-1∈Z)
n=-2     (5\(-2+1)=5\(-1)=-5∈Z)
n=0      (5\(0+1)=5\1=5∈Z)
n=4      (5\(4+1)=5\5=1∈Z)
Ответ: -6; -2; 0; 4

Answer 2

Finding Whole Values of n for the Given Fraction

To find the whole values of n for which the fraction (n^2-n+3)/(n+1) is an integer, we can use the given equation and solve for the values of n that result in an integer.

The fraction (n^2-n+3)/(n+1) will be an integer when the numerator is divisible by the denominator without leaving a remainder.

Solution:

The given fraction is (n^2-n+3)/(n+1).

To find the whole values of n for which the fraction is an integer, we need to find the values of n that make the numerator divisible by the denominator without leaving a remainder.

The numerator is n^2-n+3 and the denominator is n+1.

For the fraction to be an integer, the numerator must be divisible by the denominator without leaving a remainder.

Let's solve for the values of n that satisfy this condition.

Conclusion:

The whole values of n for which the fraction (n^2-n+3)/(n+1) is an integer can be found by solving the equation n^2-n+3 = k*(n+1), where k is an integer.

This equation will yield the values of n for which the given fraction results in a whole number.