Question

tìm x,y biết rằng 
\(x+\left(\dfrac{-31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x=y^2\)

Answer 1

\(x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x=y^2\\ \Leftrightarrow2x=y^2=\left(\dfrac{49}{12}\right)^2-\left(-\dfrac{31}{12}\right)^2\\ \Leftrightarrow2x=y^2=\left(\dfrac{49}{12}+\dfrac{31}{12}\right)\left(\dfrac{49}{12}-\dfrac{31}{12}\right)\\ \Leftrightarrow2x=y^2=\dfrac{20}{3}\cdot\dfrac{3}{2}=10\\ \Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\pm\sqrt{10}\end{matrix}\right.\)