Ta có: \(\left(2c+1\right)\left(3y-2\right)=13\)
\(\Leftrightarrow2c+1=\frac{13}{3y-2}\)
Để \(2c+1\) ∈ Z \(\Leftrightarrow\) \(3y-2\inƯ\left(13\right)\)
Mà Ư(13) = {\(\pm1;\pm13\)}
Suy ra: \(3y-2=1\Rightarrow y=1\Rightarrow c=6\left(t/m\right)\)
hoặc \(3y-2=-1\Rightarrow y=\frac{1}{3}\left(loại\right)\)
hoặc \(3y-2=13\Rightarrow y=5\Rightarrow c=0\left(t/m\right)\)
hoặc \(3y-2=-13\Rightarrow y=\frac{-11}{3}\left(loại\right)\)
Vậy các cặp (c ; y) thỏa mãn là (6 ; 1) và (0 ; 5)
(2c+1).(3y-2) = 13
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}2c+1=13\\3y-2=1\end{matrix}\right.\\\left\{{}\begin{matrix}2c+1=-13\\3y-2=-1\end{matrix}\right.\\\left\{{}\begin{matrix}2c+1=1\\3y-2=13\end{matrix}\right.\\\left\{{}\begin{matrix}2c+1=-1\\3y-2=-13\end{matrix}\right.\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}2c=13-1=12\\3y=1+2=3\end{matrix}\right.\\\left\{{}\begin{matrix}2c=-13-1=-14\\3y=-1+3=1\end{matrix}\right.\\\left\{{}\begin{matrix}2c=1-1=0\\3y=13+2=15\end{matrix}\right.\\\left\{{}\begin{matrix}2c=-1-1=-2\\3y=-13+2=-11\end{matrix}\right.\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}c=12:2=6\\y=3:3=1\end{matrix}\right.\\\left\{{}\begin{matrix}c=-14:2=-7\\y=\frac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}c=0:2=0\\y=15:3=5\end{matrix}\right.\\\left\{{}\begin{matrix}c=-2:2=-1\\y=-11:3=-\frac{11}{3}\end{matrix}\right.\end{matrix}\right.\)
Mà theo đề: c,y phải là số nguyên
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}c=6\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}c=0\\y=5\end{matrix}\right.\end{matrix}\right.\)
Vây:...........
(2c+1).(3y-2) = 13
=> ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣{2c+1=133y−2=1{2c+1=−133y−2=−1{2c+1=13y−2=13{2c+1=−13y−2=−13[{2c+1=133y−2=1{2c+1=−133y−2=−1{2c+1=13y−2=13{2c+1=−13y−2=−13
=> ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣{2c=13−1=123y=1+2=3{2c=−13−1=−143y=−1+3=1{2c=1−1=03y=13+2=15{2c=−1−1=−23y=−13+2=−11
vái copy như thế này ạ!!!
