Question

Tìm số nguyên x biết

5x+7 chia hết cho x+1

Answer 1

Ta có 5x + 7 = 5( x + 1 ) + 2

Để 5x + 7 chia hết cho x + 1

=> 5( x + 1 ) + 2 chia hết cho x + 1 

=>2 chia hết cho x + 1

=> x + 1 c Ư(2) = { 1 ; 2 ; -1 ; -2 }

=> x c { 0 ; 1 ; -2 ; -3 }

Vậy...

Answer 2

\(\frac{5x+7}{x+1}=\frac{5\left(x+1\right)+2}{x+1}=5+\frac{2}{x+1}\)

Để 5x+7\(⋮\)x+1 thì x+1\(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Xét bảng ( tự xét )

KL

Answer 3

\(5x+7⋮x+1\)

\(5\left(x+1\right)+6⋮x+1\)

Vì \(5\left(x+1\right)⋮x+1\)

\(6⋮x+1\)

\(\Rightarrow x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Ta lập bảng 

x+1	1	-1	2	-2	3	-3	6	-6
x	0	-2	1	-3	2	-4	5	-7

Answer 4

\(\left(5x+7\right)⋮x+1\)

\(=>5.\left(x+1\right)+2⋮x+1\)

Do \(5.\left(x+1\right)⋮x+1\)

\(=>2⋮x+1\)

\(=>x+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

\(=>x\in\left\{-3;-2;0;1\right\}\)

Vậy ...