\(x+\left(\frac{-31}{12}\right)^2=\left(-\frac{49}{12}\right)^2-x=y^2\)
\(\Rightarrow x+\frac{31^2}{12^2}=\frac{49^2}{12^2}-x=y^2\) (1)
\(\Rightarrow x+x=\frac{49^2}{12^2}-\frac{31^2}{12^2}\)
\(\Rightarrow2x=\frac{49^2-31^2}{12^2}\)
\(\Rightarrow2x=\frac{\left(49-31\right).\left(49+31\right)}{144}\)
\(\Rightarrow2x=\frac{18.80}{144}\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=10:2=5\)
Thay \(x=5\) vào (1) ta có:
\(5+\frac{31^2}{12^2}=y^2\)
\(\Rightarrow5+\frac{961}{144}=y^2\)
\(\Rightarrow\frac{1681}{144}=y^2\)
\(\Rightarrow\left[\begin{array}{nghiempt}y=\frac{41}{12}\\y=\frac{-41}{12}\end{array}\right.\)
Vậy \(x=5;y\in\left\{\frac{41}{12};\frac{-41}{12}\right\}\)
