Question

[Nguyễn Hữu Phúc]

   tim n thuoc Z  : 

a)n^2+1 chia het cho n+1

b)n^2-3 chia het cho n+2

c)*n+3 chia het cho n^2+2

Answer 1

[Khánh Ngọc]

a. \(\frac{n^2+1}{n+1}\in Z\)

Ta có : \(\frac{n^2+1}{n+1}=\frac{n\left(n+1\right)-n+1}{n+1}=n-1=0\)

\(\Leftrightarrow n=1\)

b. \(\frac{n^2-3}{n+2}\in Z\)

Ta có : \(\frac{n^2-3}{n+2}=\frac{n\left(n+2\right)-2n-3}{n+2}=n-\frac{2n+4-7}{n+2}=n-2-\frac{7}{n+2}\)

Để n^2 - 3 / n + 2 thuộc Z thì 7 / n + 2 thuộc Z, n thuộc Z

=> n + 2 thuộc { - 7 ; - 1 ; 1 ; 7 }

=> n thuộc { - 9 ; - 3 ; - 1 ; 5 }

Answer 2

[Ngoc Han ♪]

a ) Để \(n^2+1⋮n+1\)

mà \(n\left(n+1\right)⋮n+1\)

\(\Rightarrow n\left(n+1\right)-n^2-1⋮n+1\)

\(\Rightarrow n^2+n-n^2-1⋮n+1\)

\(\Rightarrow n-1⋮n+1\)

\(\Rightarrow n+1-2⋮n+1\)

mà \(n+1⋮n+1\)

\(\Rightarrow2⋮n+1\left(n\inℤ\right)\)

\(\Rightarrow n+1\inƯ\left(2\right)=\left\{1;-1;2-2\right\}\)

\(\Rightarrow n\in\left\{0;-2;1;-3\right\}\)

b ) \(n^2-3⋮n+2\)

mà \(n\left(n+2\right)⋮n+2\)

\(\Rightarrow n\left(n+2\right)-n^2+3⋮n+2\)

\(\Rightarrow n^2+2n-n^2+3⋮n+2\)

\(\Rightarrow2n+3⋮n+2\)

\(\Rightarrow2n+4-1⋮n+2\)

\(\Rightarrow2\left(n+2\right)-1⋮n+2\)

mà \(2\left(n+2\right)⋮n+2\)

\(\Rightarrow1⋮n+2\)

\(\Rightarrow n+2\in\left\{1;-1\right\}\)

\(\Rightarrow n\in\left\{-1;-3\right\}\)

c ) \(n+3⋮n^2+2\)

\(\Rightarrow n\left(n+3\right)⋮n^2+2\)

mà \(n^2+2⋮n^2+2\)

\(\Rightarrow n\left(n+3\right)-n^2-2⋮n^2+2\)

\(\Rightarrow n^2+3n-n^2-2⋮n^2+2\)

\(\Rightarrow3n-2⋮n^2+2\)

mà \(3\left(n+3\right)⋮n^2+2\left(n+3⋮n^2+2\right)\)

\(\Rightarrow3\left(n+3\right)-3n+2⋮n^2+2\)

\(\Rightarrow3n+9-3n+2⋮n^2+2\)

\(\Rightarrow11⋮n^2+2\left(n\in Z\right)\)

\(\Rightarrow n^2+2\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow n^2=9\)

\(\Rightarrow\orbr{\begin{cases}n=3\\n=-3\end{cases}}\)

Đối chiều đề bài , ta có \(n=-3\) thỏa mãn .