\(\frac{6n+5}{3n-2}\inℤ\Leftrightarrow6n+5⋮3n-2\)
\(\Rightarrow6n-4+9⋮3n-2\)
\(\Rightarrow2\left(3n-2\right)+9⋮3n-2\)
\(2\left(3n-2\right)⋮3n-2\)
\(\Rightarrow9⋮3n-2\)
\(\Rightarrow3n-2\inƯ\left(9\right)\)
\(\Rightarrow3n+2\in\left\{-1;1;-3;3;-9;9\right\}\)
\(\Rightarrow3n\in\left\{-3;-1;-5;1;-11;7\right\}\)
\(\Rightarrow n\in\left\{-1;\frac{-1}{3};\frac{-5}{3};\frac{1}{3};\frac{-11}{3};\frac{7}{3}\right\}\) mà n là số nguyên
\(\Rightarrow n=-1\)
\(E=\frac{6n+5}{3n-2}=\frac{6n-4+9}{3n-2}=2+\frac{9}{3n-2}\)
Để \(E\in Z\Rightarrow\frac{9}{3n-2}\in Z\)
\(\Rightarrow3n-2\inƯ\left(9\right)=\left(1;-1;3;-3;9;-9\right)\)
\(\Rightarrow3n\in\left(3;1;5;-1;11;-7\right)\)
Vì \(n\in Z\Rightarrow3n=3\Leftrightarrow n=1\)
để E là số nguyên thì\(6n+5⋮3n-2\)
Ta có : 6n+5
=6n-4+9
=2*(3n-2)+9
Mà 2*(3n-2) \(⋮\)3n-2
nên 9\(⋮\)3n-2
suy ra 3n-2 \(\in\)Ư(9)
3n-2 \(\in\){1;-1;3;-3;9;-9}
suy ra n \(\in\){\(1;\frac{1}{3};\frac{5}{3};\frac{-1}{3}\frac{11}{3};\frac{-7}{3}\)}
\(\frac{6n+5}{3n-2}=\frac{3n-2+3n+7}{3n-2}=1+\frac{3n-2+9}{3n-2}=2+\frac{9}{3n-2}\)
\(E\in Z\Leftrightarrow\frac{9}{3n-2}\in Z\Leftrightarrow9⋮3n-2\Leftrightarrow3n-2\in U\left(9\right)\)
\(\Rightarrow\)
| \(3n-2\) | 1 | -1 | 3 | -3 | 9 | -9 |
| \(n\) | 1 | \(\frac{1}{3}\left(loai\right)\) | \(\frac{5}{3}\left(loai\right)\) | \(\frac{-1}{3}\left(loai\right)\) | \(\frac{11}{3}\left(loai\right)\) | \(\frac{-7}{3}\left(loai\right)\) |
vay \(n=1\)
