Question

[Nguyễn Trịnh Diệu Linh]

Bài 1 :

Tìm x , y \(\in\) N thỏa mãn :

( 2x - 1 ) ( y - 1 ) = 29

Bài 2 :

Tìm n \(\in\) N , biết :

a) 4 n -5 \(⋮\) 2n - 1

b) 3n + 2 \(⋮\) n - 1

Bài 3 :

Cho abc \(⋮\) 7 . Chứng minh rằng :

2a + 3b + c \(⋮\) 7

Answer 1

[Nguyễn Nam]

2)

a) Ta có: \(4n-5⋮2n-1\)

\(\Rightarrow\left(4n-2\right)-3⋮2n-1\)

\(\Rightarrow2\left(2n-1\right)-3⋮2n-1\)

\(\Rightarrow-3⋮2n-1\)

\(\Rightarrow2n-1\in\left\{1;3\right\}\) ( Vì \(n\in N\) )

\(\Rightarrow\left\{{}\begin{matrix}2n-1=1\Rightarrow n=1\\2n-1=3\Rightarrow n=2\end{matrix}\right.\)

Vậy n=1 hoặc n=2

b) Ta có: \(3n+2⋮n-1\)

\(\Rightarrow\left(3n-3\right)+5⋮n-1\)

\(\Rightarrow3\left(n-1\right)+5⋮n-1\)

\(\Rightarrow5⋮n-1\)

\(\Rightarrow n-1\in\left\{1;5\right\}\) ( Vì \(n\in N\) )

\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=5\Rightarrow n=6\end{matrix}\right.\)

Vậy n=2 hoặc n=6

Answer 2

[Quang Ho Si]

1. vì (2x-1)(y-1)=29 mà \(x,y\in N\)\(\Rightarrow\left\{{}\begin{matrix}2x-1>0\\y-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\y>1\end{matrix}\right.\)

ta có:\(\left(2x-1\right)\left(y-1\right)=29\Rightarrow2x-1=\dfrac{29}{y-1}\)

vì: \(x\in N\Rightarrow\dfrac{29}{y-1}\in N\)

\(\Rightarrow29⋮y-1\Rightarrow y\in\left\{2;30\right\}\)

với y=2 => x=15

với y=30 => x=1