Question

Tìm các số thực x,y thoả mãn phương trình:

a) \(2x^2+5y^2+8x-10y+13=0\)

b)\(3x^2+5y^2-6x+20y+23=0\)

Answer 1

a) \(2x^2+5y^2+8x-10y+13=0\)

\(\Leftrightarrow\left(2x^2+8x+8\right)+\left(5y^2-10y+5\right)=0\)

\(\Leftrightarrow2\left(x+2\right)^2+5\left(y-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+2\right)^2=0\\5\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)

Vậy x=-2;y=1

b) \(3x^2+5y^2-6x+20y+23=0\)

\(\Leftrightarrow\left(3x^2-6x+3\right)+\left(5y^2+20y+20\right)=0\)

\(\Leftrightarrow3\left(x-1\right)^2+5\left(y+2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x-1\right)^2=0\\5\left(y+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy x=1;y=-2