Question

j) \(x^2+y^2+26+10x-2y=0\)

Answer 1

\(x^2+y^2+26+10x-2y=0\)

\(\left(x^2+10x\right)+\left(y^2-2y\right)+26=0\)

\(\left(x^2+2.x.5+5^2\right)+\left(y^2-2.y.1+1^2\right)=0\)

\(\left(x+5\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+5=0\\y-1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\y=1\end{array}\right.\)

Answer 2

j) \(x^2+y^2+26+10x-2y=0\)

\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow\left(x+5\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\begin{cases}x+5=0\\y-1=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=-5\\y=1\end{cases}\)

Vậy x=-5; y=1