Question

\(x+\left(-\frac{31}{12}\right)^2=\left(-\frac{49}{12}\right)^2-x=y^2\)

Tìm x và y , Giúp tớ nha . 

Answer 1

\(x+\left(\frac{-31}{12}\right)^2=\left(-\frac{49}{12}\right)^2-x=y^2\)

\(\Rightarrow x+\frac{31^2}{12^2}=\frac{49^2}{12^2}-x=y^2\) (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 \(x=5;y\in\left\{\frac{41}{12};\frac{-41}{12}\right\}\)