\(2x^2-2xy=5x-y-19\)
\(2x^2-5x+19=2xy-y\)
\(2x^2-5x+19=y\left(2x-1\right)\)
\(\dfrac{2x^2-5x+19}{2x-1}=y\)
Mà \(y\in Z\) \(\Rightarrow\dfrac{2x^2-5x+19}{2x-1}\) \(\in Z\)
Để \(\dfrac{2x^2-5x+19}{2x-1}\in Z\) \(\left(2x^2-5x+19\right)⋮\left(2x-1\right)\)
Ta có: \(\dfrac{2x^2-x-4x+2+17}{2x-1}=\dfrac{\left(2x^2-x\right)-\left(4x-2\right)+17}{2x-1}=\dfrac{x\left(2x-1\right)-2\left(2x-1\right)+17}{2x-1}\)
\(=\dfrac{\left(2x-1\right)\left(x-2\right)+17}{2x-1}=x-2+\dfrac{17}{2x-1}\)
Để \(\left(2x^2-5x+19\right)⋮\left(2x-1\right)\) thì 17 \(⋮\left(2x-1\right)\)
\(\Rightarrow\) 2x - 1 = 1; 2x - 1 = -1; 2x - 1 = 17; 2x - 1 = -17
*) 2x - 1 = 1
2x = 2
x = 1 (nhận)
*) 2x - 1 = -1
2x = 0
x = 0 (nhận)
*) 2x - 1 = 17
2x = 18
x = 9 (nhận)
*) 2x - 1 = -17
2x = -16
x = -8 (nhận)
Với x = 1 \(\Rightarrow y=\dfrac{2x^2-5x+19}{2x-1}=\dfrac{2.1^2-5.1+19}{2.1-1}=16\) (nhận) \(\Rightarrow\left(1;16\right)\)
Với x = 0 \(\Rightarrow y=\dfrac{2x^2-5x+19}{2x-1}=\dfrac{2.0^2-5.0+19}{2.0-1}=-19\) (nhận) \(\Rightarrow\left(0;19\right)\)
Với x = 9 \(\Rightarrow y=\dfrac{2x^2-5x+19}{2x-1}=\dfrac{2.9^2-5.9+19}{2.9-1}=8\) (nhận) \(\Rightarrow\left(9;8\right)\)
Với x = \(-8\) \(\Rightarrow y=\dfrac{2x^2-5x+19}{2x-1}=\dfrac{2.\left(-8\right)^2-5.\left(-8\right)+19}{2.\left(-8\right)-1}=-11\) (nhận) \(\Rightarrow\left(-8;-11\right)\)
Vậy có các cặp giá trị (x; y) sau:
(1; 16); (0; 19); (9; 8); (-8; -11)
