( x + 22 ) \(⋮\)( x + 1 )

x + 1 + 21 \(⋮\)( x + 1 )

Mà x + 1 \(⋮\)x + 1 → 21 \(⋮\)x + 1 \(\in\)Ư ( 21 )

( x - 2 ) . ( 2y + 1 ) = 17

Mà 17 là số nguyên tố và bằng 1 . 17

→ Nếu ( x - 2 ) = 1 thì ( 2y + 1 ) = 17

→ Nếu ( 2y + 1 ) = 1 thì ( x - 2 ) = 17

a, x+22 chia hết cho x+1

suy ra : x+1+21 chia hêt cho x+1

mà x+1 chia hết cho x+1 

suy ra 21 chia hết cho x+1 

suy ra x+1 thuộc -1, 1 , 3, -3, 7, -7, 21, -21

suy ra x thuộc -2, 0, 2, -4, 6, -8, 20, -22

b, 2x+23 thuộc x-1

suy ra 2x+23 = x-1

         2x-x= -23-1

         x= -24

c, 3x+1 chia hết cho 2x-1 

suy ra 2(3x+1) chia hết cho 2x-1

         6x+2 chia hết cho 2x-1           (1)

lai có 2x-1 chia hết cho 2x-1

suy ra 3(2x-1) chia hết cho 2x-1

          6x-3 chia hết cho 2x -1             (2)

từ 1 và 2

suy ra (6x+2)-(6x-3) chia hết cho 2x-1

          5 chia hết cho 2x-1 

suy ra 2x-1 thuộc -1,1,5,-5

x thuộc 0 , 1, 3, -2

c va d thì x thuộc z mới tìm được

K CHO MÌNH NHÉ

a. Ta có: n²-7 \(⋮\) n+3

<=> n²-9+2 \(⋮\) n+3

<=> (n-3)(n+3)+2\(⋮\) n+3

<=> 2 \(⋮\) n+3

=> n+3\(\in\)Ư₍₂₎=\(\left\{\pm1;\pm2\right\}\)

Ta có bảng sau:

Vậy n\(\in\left\{-2;-4;-1;-5\right\}\)

1. c, x(y - 3) = -12
Do x; y \(\in Z\Rightarrow y-3\in Z\)
Mà x(y - 13) = -12
=> x; y - 13 \(\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng :

@Đào Thị Ngọc Ánh

a, (x - 1)(y + 2) = 7
Do x; y \(\in Z\Rightarrow x-1;y+2\in Z\)
Mà (x - 1)(y + 2) = 7
=> x - 1; y + 2 \(\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Nếu \(\left\{{}\begin{matrix}x-1=1\\y+2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=-1\\y+2=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-9\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=7\\y+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=-1\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=-7\\y+2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-3\end{matrix}\right.\) (thỏa mãn)

Vậy các cặp (x; y) thỏa mãn là (2; 5); (0; -9); (8; -1); (-6; -3)
@Đào Thị Ngọc Ánh

1. b, xy - 3x - y = 0
<=> xy - 3x = y
<=> x(y - 3) = y
<=> x(y - 3) - 3 = y - 3
<=> x(y - 3) - (y - 3) = 3
<=> (x - 1)(y - 3) = 3
Do x; y \(\in Z\Rightarrow x-1;y-3\in Z\)
Mà (x - 1)(y - 3) = 3
=> x - 1; y - 3 \(\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Nếu \(\left\{{}\begin{matrix}x-1=1\\y-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=-1\\y-3=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y=0\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=3\\y-3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\) (thỏa mãn)

Nếu \(\left\{{}\begin{matrix}x-1=-3\\y-3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\) (thỏa mãn)
Vậy các cặp (x; y) thỏa mãn là (2; 6); (0; 0); (4; 4); (-2; 2)
@Đào Thị Ngọc Ánh

a) Ta có:

\(n\left(2n-3\right)-2n\left(n+1\right)\)

\(=2n^2-3n-2n^2-2n\)

\(=-5n\)

Vì \(-5n⋮5\) với n thuộc Z

\(\Rightarrow n\left(2n-3\right)-2n\left(n+1\right)⋮5\) với n thuộc Z

b) Ta có:

\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)

\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)

\(=5n^2+5n\)

\(=5\left(n^2+n\right)\)

Vì \(5\left(n^2+n\right)⋮5\)

\(\Rightarrow\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5\)

c) Ta có:

\(\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)

\(=\left(xy+1-2\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)

\(=\left(xy+1\right)\left(x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)\)

\(=\left(xy+1\right)\left(x^{2003}+y^{2003}-x^{2003}+y^{2003}\right)-2\left(x^{2003}+y^{2003}\right)\)

\(=2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)\)

Vì \(2\left(xy+1\right)y^{2003}⋮2\)

\(2\left(x^{2003}+y^{2003}\right)⋮2\)

\(\Rightarrow2\left(xy+1\right)y^{2003}-2\left(x^{2003}+y^{2003}\right)⋮2\)

\(\Rightarrow\left(xy-1\right)\left(x^{2003}+y^{2003}\right)-\left(xy+1\right)\left(x^{2003}-y^{2003}\right)⋮2\)

cái gì?

a)Ta có:

\(\left(n+5\right)⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1+6\right)⋮\left(n-1\right)\)

\(\Rightarrow6⋮\left(n-1\right)\)

Ta có bảng sau:

b)\(\left(2n-4\right)⋮\left(n+2\right)\)

\(\Rightarrow\left(2n+4-8\right)⋮\left(n+2\right)\)

\(\Rightarrow8⋮\left(n+2\right)\)

Ta có bảng sau:

c)Ta có:

\(\left(6n+4\right)⋮\left(2n+1\right)\)

\(\Rightarrow\left(6n+3+1\right)⋮\left(2n+1\right)\)

\(\Rightarrow1⋮\left(2n+1\right)\)

Ta có bảng sau:

d)Ta có:

\(\left(3-2n\right)⋮\left(n+1\right)\)

\(\Rightarrow\left(-2n-2+5\right)⋮\left(n+1\right)\)

\(\Rightarrow5⋮\left(n+1\right)\)

Ta có bảng sau:

Ta có:

\(M=5+5^2+5^3+...+5^{101}\)

\(\Rightarrow M=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{99}+5^{100}\right)+5^{101}\)

\(\Rightarrow M=30+5^3\left(1+5\right)+....+5^{99}\left(1+5\right)+5^{101}\)

\(\Rightarrow M=30+6.5^3+...+6.5^{99}+5^{101}\) có tận cùng bằng 5

⇒c=5

Mà \(\overline{abcd}⋮25\Rightarrow\overline{cd}⋮25\Rightarrow\overline{5d}⋮25\Rightarrow d=0\)

Lại có:

\(\overline{ab}=a+b^2\Rightarrow10a+b=a+b^2\)

\(\Rightarrow10a-a=b^2-b\Rightarrow9a=b\left(b-1\right)\)

\(\Rightarrow b\left(b-1\right)⋮9\)

\(\Rightarrow\left[{}\begin{matrix}b⋮9\\\left(b-1\right)⋮9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}b=9\\\varnothing\end{matrix}\right.\)

\(\Rightarrow9a=9.8=72\Rightarrow a=8\)

Vậy \(\overline{abcd}=8950\)