Question

Tìm x thuộc N biết:

3(x+4)+4(x+3) chia hết cho x+2

Answer 1

Ta có: 3(x + 4) + 4(x + 3) = 3x + 12 + 4x + 12 = 7x + 24 = 7(x + 2) + 10

Do : 7(x + 2) \(⋮\)x + 2

Để 7(x + 2) + 10 \(⋮\)x + 2 thì  10 \(⋮\)x + 2 => x + 2 \(\in\)Ư(10) = {1; 2; 5; 10}

Lập bảng :

x + 2	1	2	5	10
x	-1(loại)	0	3	8

Vậy ....

Answer 2

\(3\left(x+4\right)+4\left(x+3\right)=3x+12+4x+12\)

\(=7x+24=7x+14+10=7\left(x+2\right)+10\)

\(\Rightarrow3\left(x+4\right)+4\left(x+3\right)\)\(⋮\)\(x+2\)\(\Leftrightarrow x+2\inƯ_{10}\)

Mà \(Ư_{10}=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)\(\Rightarrow...\)

Answer 3

\(3\left(x+4\right)+4\left(x+3\right)⋮x+2\)

\(\Rightarrow3x+12+4x+12⋮x+2\)

\(\Rightarrow7x+24⋮x+2\)

\(\Rightarrow7x+14+10⋮x+2\)

\(\Rightarrow2\left(x+2\right)+10⋮x+2\)

Ta có:\(2\left(x+2\right)⋮x+2\Rightarrow10⋮x+2\)

\(\Rightarrow x+2\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

Mà \(x\inℕ\Rightarrow x+2\ge2\Rightarrow x+2\in\left\{2;5;10\right\}\)

\(x\in\left\{0;3;7\right\}\)

Vậy\(x=0;3;7\)

Answer 4

Ta có : 3(x + 4) + 4(x + 3) \(⋮\)x + 2

=> 3x + 12 + 4x + 12 \(⋮\)x + 2

=> 3x + 4x + 12 + 12 \(⋮\)x + 2

=> 7x + 24 \(⋮\)x + 2

=> 7x + 14 + 10 \(⋮\)x + 2

=> 7(x + 2) + 10 \(⋮\)x + 2

Ta có : Vì \(7\left(x+2\right)⋮x+2\)

          => \(10⋮x+2\)

          => \(x+2\inƯ\left(10\right)\)

          => \(x+2\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

Lập bảng xét các trường hợp ta có :

\(x+2\)	\(1\)	\(-1\)	\(2\)	\(-2\)	\(5\)	\(-5\)	\(10\)	\(-10\)
\(x\)	\(-1\)	\(-3\)	\(0\)	\(-4\)	\(3\)	\(-7\)	\(8\)	\(-12\)

Vậy \(3\left(x+4\right)+4\left(x+3\right)⋮x+2\)\(\Leftrightarrow x\in\left\{-1;-3;0;-4;3;-7;8;-12\right\}\)