\(y+2⋮x;x+2⋮y\Rightarrow\left(x+2\right)\left(y+2\right)⋮xy\Rightarrow xy+2x+2y+4⋮xy\Rightarrow2x+2y+4⋮xy\)

\(\Rightarrow2\left(x+y+2\right)⋮xy\Rightarrow2⋮xy\Rightarrow xy\inƯ\left(2\right)=1;2\)

\(xy=1\Rightarrow x=1,y=1\Rightarrow y+2=1+2=3⋮x=1\Rightarrow y+2⋮x\)

                                             \(x+2=1+2=3⋮y=1\Rightarrow x+2⋮y\)

\(\Rightarrow x=1,y=1\left(tm\right)\)

\(xy=2\Rightarrow x=1,y=2;x=2,y=1\Rightarrow x+2=1+2=3\)ko chia hết cho \(y=2\Rightarrow x+2\)ko chia hết cho y

\(\Rightarrow x=1,y=2\left(ktm\right)\Rightarrow x=2,y=1\left(ktm\right)\)

vậy x=1,y=1 

ko chắc lắm

yeah yeah

6   \(n^5+5n=n^5-n+6n=n\left(n^4-1\right)+6n=n\left(n^2-1\right)\left(n^2+1\right)+6n\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+6n\)

vì n,n-1 là 2 số nguyên lien tiếp  \(\Rightarrow n\left(n-1\right)⋮2\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\)

  n,n-1,n+1 là 3 sô nguyên liên tiếp \(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮3\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮3\)

\(\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2\cdot3=6\)

\(6⋮6\Rightarrow6n⋮6\Rightarrow n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)-6n⋮6\Rightarrow n^5+5n⋮6\)(đpcm)

7   \(n\left(2n+7\right)\left(7n+1\right)=n\left(2n+7\right)\left(7n+7-6\right)=7n\left(n+1\right)\left(2n+7\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4+3\right)-6n\left(2n+7\right)\)

\(=7n\left(n+1\right)\left(2n+4\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

\(=14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)\)

n,n+1,n+2 là 3 sô nguyên liên tiếp dựa vào bài 6 \(\Rightarrow n\left(n+1\right)\left(n+2\right)⋮6\Rightarrow14n\left(n+1\right)\left(n+2\right)⋮6\)

\(21⋮3;n\left(n+1\right)⋮2\Rightarrow21n\left(n+1\right)⋮3\cdot2=6\)

\(6⋮6\Rightarrow6n\left(2n+7\right)⋮6\)

\(\Rightarrow14n\left(n+1\right)\left(n+2\right)+21n\left(n+1\right)-6n\left(2n+7\right)⋮6\)

\(\Rightarrow n\left(2n+7\right)\left(7n+1\right)⋮6\)(đpcm)

......................?

mik ko biết

mong bn thông cảm 

nha ................

\(\left(n+5\right)⋮\left(n+1\right)\)

\(\Rightarrow\left(n+1\right)+4⋮\left(n+1\right)\)

\(\Rightarrow4⋮\left(n+1\right)\Rightarrow\left(n+1\right)\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Do \(n\in N\)

\(\Rightarrow n\in\left\{0;1;3\right\}\)

(n+5)⋮(n+1)(n+5)⋮(n+1)

⇒(n+1)+4⋮(n+1)⇒(n+1)+4⋮(n+1)

⇒4⋮(n+1)⇒(n+1)∈Ư(4)={1;−1;2;−2;4;−4}⇒4⋮(n+1)⇒(n+1)∈Ư(4)={1;−1;2;−2;4;−4}

Do 

Vì vai trò của x, y bình đẳng nên có thể giả sử 
x
≤
y
x≤y.
- Nếu x = 1 thì 
x
+
1
=
2
⋮
y
x+1=2⋮y 
⇒
y
=
1
⇒y=1 hoặc 2 
⇒
(
x
,
y
)
=
(
1
,
1
)
,
(
1
,
2
)
⇒(x,y)=(1,1),(1,2).
 
- Nếu 
x
≥
2
x≥2 thì 
2
≤
x
≤
y
2≤x≤y
Có 
⎧
⎨
⎩
x
+
1
⋮
y
y
+
1
⋮
x
{x+1⋮yy+1⋮x
⇒
(
x
+
1
)
(
y
+
1
)
=
(
x
y
+
x
+
y
+
1
)
⋮
x
y
⇒(x+1)(y+1)=(xy+x+y+1)⋮xy 
⇒
(
x
+
y
+
1
)
⋮
x
y
⇒(x+y+1)⋮xy
⇒
x
+
y
+
1
x
y
=
1
x
+
1
y
+
1
x
y
⇒x+y+1xy=1x+1y+1xy là số nguyên dương.
Mà 
2
≤
x
≤
y
2≤x≤y nên 
1
x
+
1
y
+
1
x
y
≤
1
2
+
1
2
+
1
4
=
5
4
1x+1y+1xy≤12+12+14=54
Từ đó suy ra 
1
x
+
1
y
+
1
x
y
=
1
1x+1y+1xy=1 (1)
⇒
1
=
1
x
+
1
y
+
1
x
y
≤
1
x
+
1
x
+
1
2
x
=
5
2
x
⇒1=1x+1y+1xy≤1x+1x+12x=52x 
⇒
2
x
≤
5
⇒2x≤5 
⇒
⇒ x = 2
Thay vào (1) ta có 
1
2
+
1
y
+
1
2
y
=
1
12+1y+12y=1 
⇒
y
=
3
⇒y=3
 
Vậy các cặp số (x, y) phải tìm là (1, 1), (1, 2), (2, 1), (2, 3), (3, 2).

mình lớp 5 có gì mong bạn thông cảm

\(1,\left(x-1\right)\left(y-2\right)=2\left(x,y\in N\right)\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=2\\y-2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=1\\y-2=2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left(x;y\right)=\left\{\left(3;3\right);\left(2;4\right)\right\}\)