Đề bài bài này là : Tìm n thuộc N, biết
a) Ta có:
\(4n+7⋮3n-2\)
\(\Rightarrow3\left(4n+7\right)⋮3n-2\)
\(\Rightarrow12n+21⋮3n-2\)
\(\Rightarrow\left(12n-8\right)+29⋮3n-2\)
\(\Rightarrow4\left(3n-2\right)+29⋮3n-2\)
\(\Rightarrow29⋮3n-2\)
\(\Rightarrow3n-2\in U\left(29\right)=\left\{1;29\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}3n-2=1\Rightarrow n=1\\3n-2=29\Rightarrow n=\dfrac{31}{3}\left(loai\right)\end{matrix}\right.\)
Vậy \(n=1\)
b) Ta có:
\(6n+8⋮5n-2\)
\(\Rightarrow5\left(6n+8\right)⋮5n-2\)
\(\Rightarrow30n+40⋮5n-2\)
\(\Rightarrow\left(30n-12\right)+52⋮5n-2\)
\(\Rightarrow6\left(5n-2\right)+52⋮5n-2\)
\(\Rightarrow52⋮5n-2\)
\(\Rightarrow5n-2\in U\left(52\right)=\left\{1;2;4;13;26;52\right\}\) ( Vì \(n\in N\) )
+) \(5n-2=1\Rightarrow n=\dfrac{3}{5}\left(loai\right)\)
+) \(5n-2=2\Rightarrow n=\dfrac{4}{5}\left(loai\right)\)
+) \(5n-2=4\Rightarrow n=\dfrac{6}{5}\left(loai\right)\)
+) \(5n-2=13\Rightarrow n=3\left(thoa\right)\)
+) \(5n-2=26\Rightarrow n=\dfrac{28}{5}\left(loai\right)\)
+) \(5n-2=52\Rightarrow n=\dfrac{54}{5}\left(loai\right)\)
Vậy \(n=3\)
c) Ta có:
\(2n+1⋮3n-4\)
\(\Rightarrow3\left(2n+1\right)⋮3n-4\)
\(\Rightarrow6n+3⋮3n-4\)
\(\Rightarrow\left(6n-8\right)+11⋮3n-4\)
\(\Rightarrow2\left(3n-4\right)+11⋮3n-4\)
\(\Rightarrow11⋮3n-4\)
\(\Rightarrow3n-4\in U\left(11\right)=\left\{1;11\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}3n-4=1\Rightarrow n=\dfrac{5}{3}\left(loai\right)\\3n-4=11\Rightarrow n=5\end{matrix}\right.\)
Vậy \(n=5\)
