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d) Để \(\dfrac{n+1}{2n+1}\in Z\) thì \(n+1⋮2n+1\)
\(\Leftrightarrow1⋮2n+1\)
\(\Leftrightarrow2n+1\in\left\{1;-1\right\}\)
\(\Leftrightarrow2n\in\left\{0;-2\right\}\)
hay \(n\in\left\{0;-1\right\}\)
Mk trả lời mỗi câu khó nha!!!
d*) \(\dfrac{n+1}{2n+1}\in Z\)
Để \(\dfrac{n+1}{2n+1}\in Z\) thì \(n+1⋮2n+1\)
\(n+1⋮2n+1\)
\(\Rightarrow2.\left(n+1\right)⋮2n+1\)
\(\Rightarrow2n+2⋮2n+1\)
\(\Rightarrow2n+1+1⋮2n+1\)
\(\Rightarrow1⋮2n+1\)
\(\Rightarrow2n+1\inƯ\left(1\right)=\left\{\pm1\right\}\)
Ta có bảng giá trị:
| 2n+1 | -1 | 1 |
| n | -1 | 0 |
Vậy \(n\in\left\{-1;0\right\}\)
a: \(\Leftrightarrow2n-1\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{1;0;2;-1\right\}\)
c: \(\Leftrightarrow n+1\in\left\{1;-1\right\}\)
hay \(n\in\left\{0;-2\right\}\)
a, \(3n+2⋮n-1\)
\(\Rightarrow3n-3+5⋮n-1\)
\(\Rightarrow3\left(n-1\right)+5⋮n-1\)
Vì : \(3\left(n-1\right)⋮n-1\Rightarrow5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(5\right)\)
\(\Rightarrow n-1\in\left\{1;5\right\}\)
+) \(n-1=1\Rightarrow n=1+1\Rightarrow n=2\)
+) \(n-1=5\Rightarrow n=5+1\Rightarrow n=6\)
Vậy : \(n\in\left\{2;6\right\}\) thì \(3n+2⋮n-1\)
b, \(n+8⋮n+3\)
Vì : \(n+3⋮n+3\)
\(\Rightarrow\left(n+8\right)-\left(n+3\right)⋮n+3\)
\(\Rightarrow n+8-n-3⋮n+3\)
\(\Rightarrow5⋮n+3\)
\(\Rightarrow n+3\inƯ\left(5\right)\)
Mà : \(n+3\ge3\)
\(\Rightarrow n+3=5\Rightarrow n=5-3\Rightarrow n=2\)
Vậy n = 2 thì : \(n+8⋮n+3\)
c, \(n+6⋮n-1\)
Mà : \(n-1⋮n-1\)
\(\Rightarrow\left(n+6\right)-\left(n-1\right)⋮n-1\)
\(\Rightarrow n+6-n+1⋮n-1\)
\(\Rightarrow7⋮n-1\)
\(\Rightarrow n-1\inƯ\left(7\right)\)
\(\Rightarrow n-1\in\left\{1;7\right\}\)
+) \(n-1=1\Rightarrow n=1+1\Rightarrow n=2\)
+) \(n-1=7\Rightarrow n=7+1\Rightarrow n=8\)
Vậy \(n\in\left\{2;8\right\}\) thì \(n+6⋮n-1\)
d, \(4n-5⋮2n-1\)
\(\Rightarrow4n-2-3⋮2n-1\)
\(\Rightarrow2\left(2n-1\right)-3⋮2n-1\)
Vì : \(2\left(2n-1\right)⋮2n-1\)
\(\Rightarrow3⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(3\right)\)
\(\Rightarrow2n-1\in\left\{1;3\right\}\)
+) \(2n-1=1\Rightarrow2n=1+1\Rightarrow2n=2\Rightarrow n=2\div2\Rightarrow n=1\)
+) \(2n-1=3\Rightarrow2n=3+1\Rightarrow2n=4\Rightarrow n=4\div2\Rightarrow n=2\)
Vậy \(n\in\left\{1;2\right\}\) thì \(4n-5⋮2n-1\)
a) 2n - 4 ⋮ n - 3
2n - 6 + 2 ⋮ n - 3
2( n - 3 ) + 2 ⋮ n - 3
Vì 2( n - 3 ) ⋮ n - 3
=> 2 ⋮ n - 3
=> n - 3 thuộc Ư(2) = { 1; -1; 2; -2 }
=> n thuộc { 4; 2; 5; 1 }
Vậy,......
- Các câu còn lại tương tự
\(a,2n-4⋮n-3\Leftrightarrow2n-6+2⋮n-3\)
\(\Leftrightarrow2\left(n-3\right)+2⋮n-3\Leftrightarrow2⋮n-3\left(n-3\inℤ\right)\)
\(\Leftrightarrow n-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)\(\Leftrightarrow n\in\left\{2;4;1;5\right\}\)
Vậy \(n=1;2;4;5\)
a, 3n+2 chia hết n-1
=> 3(n-1)+5 chia hết cho n-1
Mà 3(n-1) chia hết cho n-1
=> 5 chia hết cho n-1
Lại có n thuộc N
=> n-1 thuộc Ư(5)=1,-1,5,-5
=> n=2,0,6,-4
a) 3n+2 chia hết cho n-1
=>3(n-1)+5 chia hết cho n-1 = 5 chia hết cho n-1
=>n-1 thuộc Ư(5)={-1;1;-5;5}
n-1=-1=>n=0 = n-1=1=>n=2
n-1=-5=>n=-4 = n-1=5=>n=6
Ta co 3×n+2 chia het n-1
Suy ra 3(x-1)+5chia het x-1
Vi 3(x-1)chia het cho x-1
Suy ra 5chia het x-1
Thi x-1thuoc uoc cua 5=1,5
X thuoc 6,2