Question

Tìm x thoa mãn:

x² + y² + 26 + 10x + 2y =0

5x²+ y² - 2xy - 4x + 1 =0

2x² + 2xy - 4x - 2y + 2 + y² =0

Answer 1

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

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

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

\(\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)( do \(\left(x+5\right)^2\ge0;\left(y+1\right)^2\ge0\))

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

Answer 2

ban kia lam dung roi do

k tui nha

thanks