tim x thuoc N , biet :
a)4n=256
b)620 . 64n=6200
Tim n thuoc Z biet: 4n+7/2n-3
tim n thuoc z biet 4n-1 chia het cho 2n+3
4n -1 chia hết cho 2n-3
2n - 3 chia hết cho 2n -3
=> 2(2n-3) chia hết cho 2n - 3
=> 4n - 6 chia hết cho 2n -3
=> 4n -1- ( 4n -6) chia hết cho 2n - 3
=> 4n -1 - 4n = 6 chia hết cho 2n - 3
=> 5 chia hết cho 2n-3
=> 2n -3 thuộc ước của 5
đến đây dễ rồi bạn tự làm nhé
tim x thuoc Z biet
a)4n-5 chia het cho n
b)-11 la boi cua n-1
c)2n-1 la uoc cua 3n +2
Tìm số nguyên x,y biết
(x+1)*(y+1)=-13
\(\left(4n-5\right)⋮n\)
\(\Rightarrow5⋮n\)
\(\Rightarrow n\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow n\in\left\{\pm5;\pm1\right\}\)
\(-11\text{ là bội của }n-1\)
\(\Rightarrow n-1\inƯ\left(-11\right)=\left\{\pm1;\pm11\right\}\)
\(\Rightarrow n\in\left\{2;0';12;-10\right\}\)
tim x thuoc Z biet
a)4n-5 chia het cho n
b)-11 la boi cua n-1
c)2n-1 la uoc cua 3n +2
Tim n thuoc so tu hien biet
6n+9 chia het cho 4n -1
thuc hien phep chia ra so du bao nhieu rtoi cho no =0 giai ra n
Tim gia tri lon nhat cua phan so M=6n-3/4n-6 khi do n=? biet n thuoc N
tim a,b.c biet a thuoc n, b thuoc n, c thuoc n:a nhan bc = 145
Tim a thuoc N , biet
a) a mu x = 1 (x thuoc N*)
b) ax = 0 ( x thuoc Nx)
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 \(n=0\)