6n+9\(⋮\)4n-1 ->4.(6n+9)\(⋮\)4n-1
->24n+36\(⋮\)4n-1
->24n-6+42\(⋮\)4n-1
->6(4n-1)+42\(⋮\)4n-1
->4n-1 thuoc uoc cua 42 ma n\(\supseteq\)1 nen 4n-1\(\supseteq\)3
4n-1 | 3 | 6 | 7 | 21 | 42 |
n | 1 | 7/4 | 2 | 11/2 | 43/4 |
ma n laf so tu nhien nen n=1,2
6n+9\(⋮\)4n-1 ->4.(6n+9)\(⋮\)4n-1
->24n+36\(⋮\)4n-1
->24n-6+42\(⋮\)4n-1
->6(4n-1)+42\(⋮\)4n-1
->4n-1 thuoc uoc cua 42 ma n\(\supseteq\)1 nen 4n-1\(\supseteq\)3
4n-1 | 3 | 6 | 7 | 21 | 42 |
n | 1 | 7/4 | 2 | 11/2 | 43/4 |
ma n laf so tu nhien nen n=1,2
\(D=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{1}{\left(2n-1\right)^2}\right),\)với \(n\in N,n\ge1\)
Tìm \(n\in N\)sao cho :
a) 15 - 4n chia hết cho n
b) ( 6n - 9 ) chia hết cho n \(\left(n\ge2\right)\)
c) ( n + 13 ) chia hết cho ( n - 5 )
d) ( 15 - 2n ) chia hết cho n + 1 \(\left(n\le7\right)\)
Tìm tất cả các số nguyên tố p có dạng \(\dfrac{n\left(n+1\right)}{2}-1\left(n\ge1\right)\)
\(\frac{\left(\frac{-2}{11}\right)^{n+1}}{\left(\frac{-2}{11}\right)^n}\left(n\ge1\right)\)
tìm số nguyên n để :
a,\(\left(n+5\right)⋮\left(n+1\right)\)
b,\(\left(6n+4\right)⋮\left(2n+1\right)\)
Bài 1 : Tìm \(n\in N\)
a) \(\frac{4n-1}{3n+2}\in N\) b) \(\frac{5n-7}{2n+1}\in N\)
Bài 2 : Tìm \(n\in N\)
a) \(\left(n+2\right)\cdot\left(2n+5\right)=21\) b) \(\left(2n-3\right)\cdot\left(n-5\right)=22\)
Bài 3 : Tìm \(x.y\in N\)
a) \(\left(2n+1\right)\cdot\left(3y-5\right)=12\) b) \(\left(3x-1\right)\cdot\left(4y+3\right)=14\)
Cách bạn giải ra giúp mình nha !
Tìm số tự nhiên n, biết :
a/ \(\left(2.n-1\right)^4:\left(2.n-1\right)=27\)
b/ \(\left(2n+1\right)^5:\left(2.n+1\right)^2=1\)
c/ \(\left(n+1\right)^3:\left(n+1\right)=4\)
d/ \(\left(21+n\right):9=9^5:9^4\)
1) Cho tổng:
A = 4n + 4 \(\left(n\in Z\right)\) . Tìm n để A chia hết cho n
B = 5n + 6 \(\left(n\in Z\right)\) . Tìm n để B chia hết cho n
2) Tính nhanh
a) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}\)
b) \(\frac{1}{7}.\frac{5}{9}+\frac{5}{9}.\frac{1}{7}+\frac{5}{9}.\frac{3}{7}\)
tìm : \(n\in N\)để :
a)\(\left(2n+9\right)⋮\left(3n+1\right)\)
b)\(\left(5n+2\right)⋮\left(9-2n\right)\)