Question

Tìm x, y biết

\(29+\left(x-256\right)^2=4+y^2\)

Answer 1

a.

Ta có

\(29+\left(x-256\right)^2=4+y^2\)

\(\left(x-256\right)^2=y^2-25\)

\(\left(x-256\right)^2=\left(y-5\right).\left(y+5\right)\)

\(\left(x-256\right)^2-\left(y-5\right).\left(y+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-256=0\\y-5=0\end{matrix}\right.\\\left[{}\begin{matrix}x-256=0\\x+5=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=256\\y=5\end{matrix}\right.\\\left[{}\begin{matrix}x=256\\y=-5\end{matrix}\right.\end{matrix}\right.\)