tim n thuoc N sao cho n+3 chia het cho n-1
tim n thuoc N sao cho (3n+1)chia het cho (2n+3)
tim n thuoc Z sao cho 2n-3 chia het cho n+1
ta có n+1:n+1
2(n+1):n+1
2n+2:n+1
mà 2n-3:n+1
=)2n+2-5:n+1
n+1 thuộc Ư(5)={1;-1;5;-5}
vậy n={0;-2;4;6}
đung n
tim n thuoc N* biet:
a, n^5+1 chia het cho n^3+1
b,n^3-n chia het cho n-3
c,n^3-3 chia het cho n-3
1.chung minh rang:3n.(n+1)chia het cho 6(n thuoc N
2.cmr 5n.(n+1).(n+2) chia het cho 30(n thuocN)
3.tim so tu nhien n de 7.(n-1) chia het cho 4
4.tim so tu nhien n de 5.( n-2) chia het cho 3
Tim n thuoc N sao cho 4n-5 chia het cho 2n-1
Ta có
4n - 5 chia hết cho 2n - 1 => mà 2n - 1 cũng chia hết cho 2n - 1
=> 2( 2n - 1 ) sẽ chia hết cho 2n - 1
=> 4n - 2 chia hết cho 2n - 1 , 4n - 5 cũng chia hết cho 2n -1 => (4n - 2) - (4n - 5) chia hết cho 2n - 1
=> 3 chia hết cho 2n - 1 => 2n - 1 \( \in\) ước của 3
+) 2n - 1 = -3 => n = -1 ( loại) vì n thuộc N
+) 2n - 1 = -1 => n = 0 (ok)
+) 2n - 1 = 1 => n = 1 (ok)
+) 2n - 1 = 3 => n = 2 (ok)
vậy với n = 0; n = 1 ; n = 2 thì 4n - 5 chia hết cho 2n -1
Giải:
Ta có:
\(4n-5⋮2n-1\)
\(\Rightarrow\left(4n-2\right)-3⋮2n-1\)
\(\Rightarrow2\left(2n-1\right)-3⋮2n-1\)
\(\Rightarrow-3⋮2n-1\)
Mà n thuộc N nên \(2n-1\in\left\{1;3\right\}\)
+) \(2n-1=1\Rightarrow n=1\)
+) \(2n-1=3\Rightarrow n=2\)
Vậy \(n\in\left\{1;2\right\}\)
1/tim n thuoc N sao cho:
a/(2n+12) chia het cho (n+2)
b/(3n+5) chia het cho (n-2)
2/ tim x sao cho:
a/(x+3).(x^2+1)=0
b/(x+7).(x^2-36)=0
a/ \(2n+12⋮n+2\)
Mà \(n+2⋮n+2\)
\(\Leftrightarrow\hept{\begin{cases}2n+12⋮n+2\\2n+4⋮n+2\end{cases}}\)
\(\Leftrightarrow8⋮n+2\)
\(\Leftrightarrow n+2\inƯ\left(8\right)\)
Suy ra :
+) n + 2 = 1 => n = -1 (loại)
+) n + 2 = 2 => n = 0
+) n + 2 = 4 => n = 2
+) n + 2 = 8 => n = 6
Vậy ......
b/ \(3n+5⋮n-2\)
Mà \(n-2⋮n-2\)
\(\Leftrightarrow\hept{\begin{cases}3n+5⋮n-2\\3n-6⋮n-2\end{cases}}\)
\(\Leftrightarrow11⋮n+2\)
\(\Leftrightarrow n+2\inƯ\left(11\right)\)
\(\Leftrightarrow\orbr{\begin{cases}n+2=1\\n+2=11\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}n=-1\left(loại\right)\\n=9\end{cases}}\)
Vậy ..
a/ \(\left(x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+1=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-3\\x^2=-1\left(loại\right)\end{cases}}\)
Vậy ....
b/ \(\left(x+7\right)\left(x^2-36\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+7=0\\x^2-36=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-7\\x^2=36\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-7\\x=6or=-6\end{cases}}\)
Vậy ...
Tim n thuoc N sao cho:4n-5 chia het cho 2n-1
tim n thuoc Z
a)n^2+4chia het cho n-1
b)3n-1 chia het cho 2-n
c)n-7 chia het cho 2n+3
phần c
\(n-7⋮2n+3\)
\(2\left(n-7\right)-\left(2n+3\right)⋮2n+3\)
\(2n-4-2n-3⋮2n+3\)
\(-7⋮2n+3\)
\(\Rightarrow2n+3\inƯ\left(-7\right)=\left\{\pm1;\pm7\right\}\)
Ta có bảng xét :
2n+3 | -1 | 1 | -7 | 7 |
2n | -4 | -2 | -10 | 4 |
n | -1 | 1 | -5 | 2 |
Tim so tu nhien n sao cho:
a/ 5:n+1 b/ 15:n+1 c/ n+3 : n+1 d/ 4n+3:2n+1
Biet rang 7a+2b chia het cho 13 ( a,b thuoc N ). Chung to rang 10a+b cung chia het cho 13 ?
a) Ta có:
\(5⋮n+1\)
\(\Rightarrow n+1\in U\left(5\right)=\left\{1;5\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=5\Rightarrow n=4\end{matrix}\right.\)
Vậy \(n\in\left\{0;4\right\}\)
b) Ta có:
\(15⋮n+1\)
\(\Rightarrow n+1\in U\left(15\right)=\left\{1;3;5;15\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=3\Rightarrow n=2\\n+1=5\Rightarrow n=4\\n+1=15\Rightarrow n=14\end{matrix}\right.\)
Vậy \(n\in\left\{0;2;4;14\right\}\)
c) Ta có:
\(n+3⋮n+1\)
\(\Rightarrow\left(n+1\right)+2⋮n+1\)
\(\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\in U\left(2\right)=\left\{1;2\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=2\Rightarrow n=1\end{matrix}\right.\)
Vậy \(n\in\left\{0;1\right\}\)
d) Ta có:
\(4n+3⋮2n+1\)
\(\Rightarrow\left(4n+2\right)+1⋮2n+1\)
\(\Rightarrow2\left(2n+1\right)+1⋮2n+1\)
\(\Rightarrow1⋮2n+1\)
\(\Rightarrow2n+1\in U\left(1\right)=\left\{1\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow2n+1=1\)
\(\Rightarrow n=0\)
Vậy \(n=0\)