Question

Tìm số nguyên n biết 

1) 3n+7 chia hết cho 2 n+1

2) n²+25 chia hết cho n+2

3) 3n²+5 chia hết cho n-1

Ai là đúng nhất thì mình sẽ tịck cho

Answer 1

1) Để \(3n+7⋮2n+1\) \(\Leftrightarrow\)\(2.\left(3n+7\right)⋮2n+1\)

- Ta có: \(2.\left(3n+7\right)=6n+14=\left(6n+3\right)+11=3.\left(2n+1\right)+11\)

-  Để \(2.\left(3n+7\right)⋮2n+1\)\(\Rightarrow\)\(3.\left(2n+1\right)+11⋮2n+1\)mà \(3.\left(2n+1\right)⋮2n+1\)

\(\Rightarrow\)\(11⋮2n+1\)\(\Rightarrow\)\(2n+1\inƯ\left(11\right)\in\left\{\pm1;\pm11\right\}\)

- Ta có bảng giá trị:

\(2n+1\)	\(-1\)	\(1\)	\(-11\)	\(11\)
\(n\)	\(-1\)	\(0\)	\(-6\)	\(5\)
	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)

Vậy \(n\in\left\{-6,-1,0,5\right\}\)

2) Ta có: \(n^2+25=\left(n^2-4\right)+29=\left(n+2\right).\left(n-2\right)+29\)

- Để \(n^2+25⋮n+2\)\(\Rightarrow\)\(\left(n+2\right).\left(n-2\right)+29⋮n+2\)mà \(\left(n+2\right).\left(n-2\right)⋮n+2\)

\(\Rightarrow\)\(29⋮n+2\)\(\Rightarrow n+2\inƯ\left(29\right)\in\left\{\pm1;\pm29\right\}\)

- Ta có bảng giá trị: 

\(n+2\)	\(-1\)	\(1\)	\(-29\)	\(29\)
\(n\)	\(-3\)	\(-1\)	\(-31\)	\(27\)
	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)

 Vậy \(n\in\left\{-31,-3,-1,27\right\}\)

3) Ta có: \(3n^2+5=\left(3n^2-3\right)+8=3.\left(n+1\right).\left(n-1\right)+8\)

- Để \(3n^2+5⋮n-1\)\(\Rightarrow\)\(3.\left(n+1\right).\left(n-1\right)+8⋮n-1\)mà \(3.\left(n+1\right).\left(n-1\right)⋮n-1\)

\(\Rightarrow\)\(8⋮n-1\)\(\Rightarrow n-1\inƯ\left(8\right)\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

- Ta có bảng giá trị:

\(n-1\)	\(-1\)	\(1\)	\(-2\)	\(2\)	\(-4\)	\(4\)	\(-8\)	\(8\)
\(n\)	\(0\)	\(2\)	\(-1\)	\(3\)	\(-3\)	\(5\)	\(-7\)	\(9\)
	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)

Vậy \(n\in\left\{-7,-3,-1,0,2,3,5,9\right\}\)