Question

tìm số nguyên dương n sao cho:

a) (n+1)(n+2) chia hết cho n;

b) (n+2)(n+3) chia hết cho n;

c) (n-1)(n+3) chia hết cho n

 

Answer 1

to be continued ._.

Answer 2

a,                                                                      Bài giải

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

\(=\frac{n\left(n+1+2\right)}{n}+\frac{2}{n}=n+1+2+\frac{2}{n}\)

\(\left(n+1\right)\left(n+2\right)\text{ }⋮\text{ }n\text{ khi }2\text{ }⋮\text{ }n\)

\(\Rightarrow\text{ }n\inƯ\left(2\right)=\left\{\pm1\text{ ; }\pm2\right\}\)

Answer 3

b,                                                         Bài giải

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

\(=\frac{n\left(n+2+3\right)}{n}+\frac{2}{n}=n+2+3+\frac{2}{n}\)

 \(\left(n+2\right)\left(n+3\right)\text{ }⋮\text{ }n\text{ khi }2\text{ }⋮\text{ }n\)

\(\Rightarrow\text{ }n\inƯ\left(2\right)=\left\{\pm1\text{ ; }\pm2\right\}\)

Answer 4

a,                                                                      Bài giải

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

\(=\frac{n\left(n+1+2\right)}{n}+\frac{2}{n}=n+1+2+\frac{2}{n}\)

\(\left(n+1\right)\left(n+2\right)\text{ }⋮\text{ }n\text{ khi }2\text{ }⋮\text{ }n\)

\(\Rightarrow\text{ }n\inƯ\left(2\right)=\left\{\pm1\text{ ; }\pm2\right\}\)

Answer 5

Sorry ! Câu ba mình làm nhầm ! Đây mình sửa lại nè :

b,                                                                      Bài giải

Ta có : \(\frac{\left(n+2\right)\left(n+3\right)}{n}=\frac{n\left(n+2\right)+3\left(n+2\right)}{n}=\frac{n^2+2n+3n+6}{n}=\frac{n\left(n+2+3\right)+6}{n}\)

\(=\frac{n\left(n+1+2\right)}{n}+\frac{6}{n}=n+1+2+\frac{6}{n}\)

\(\left(n+2\right)\left(n+3\right)\text{ }⋮\text{ }n\text{ khi }6\text{ }⋮\text{ }n\)

\(\Rightarrow\text{ }n\inƯ\left(2\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)

Answer 6

b,                                                                      Bài giải

Ta có : \(\frac{\left(n+2\right)\left(n+3\right)}{n}=\frac{n\left(n+2\right)+3\left(n+2\right)}{n}=\frac{n^2+2n+3n+6}{n}=\frac{n\left(n+2+3\right)+6}{n}\)

\(=\frac{n\left(n+1+2\right)}{n}+\frac{6}{n}=n+1+2+\frac{6}{n}\)

\(\left(n+2\right)\left(n+3\right)\text{ }⋮\text{ }n\text{ khi }6\text{ }⋮\text{ }n\)

\(\Rightarrow\text{ }n\inƯ\left(2\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)