Question

giải pt sau

\(x^2\)-2x+y\(^2\)-8y+17=0

Answer 1

\(x^2-2x+y^2-8y+17=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-8y+16\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-4\right)^2=0\)

Vì \(\left(x-1\right)^2\ge0\)

\(\left(y-4\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2=\left(y-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)

Vậy phương trình có nghiệm \(\left(x,y\right)=\left(1;4\right)\)