Ta có: \(4x^2-2xy-2x=y-20\)
\(\Leftrightarrow y+2xy=4x^2-2x+20\)
\(\Leftrightarrow y\cdot\left(2x+1\right)=4x^2-2x+20\)
\(\Leftrightarrow y=\dfrac{4x^2-2x+20}{2x+1}\)
\(\Leftrightarrow y=\dfrac{4x^2+2x-4x+20}{2x+1}\)
\(\Leftrightarrow y=\dfrac{2x\left(2x+1\right)-4x-2+22}{2x+1}\)
\(\Leftrightarrow y=2x+\dfrac{-2\left(2x+1\right)+22}{2x+1}\)
\(\Leftrightarrow y=2x-2+\dfrac{22}{2x+1}\)
Để x,y ∈ Z thì \(\dfrac{22}{2x+1}\) có giá trị nguyên
\(\Rightarrow2x+1\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
Mà nếu x nguyên thì \(2x+1\) luôn là số lẻ
\(\Rightarrow2x+1\in\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{0;-1;5;-6\right\}\)
Ta tìm được các số y tương ứng là:
\(x=0\Rightarrow y=20\)
\(x=-1\Rightarrow y=-26\)
\(x=5\Rightarrow y=10\)
\(x=-6\Rightarrow y=-16\)
Vậy các cặp x,y thỏa là: \(\left(0;20\right);\left(-1;-26\right);\left(5;10\right);\left(-6;-16\right)\)
