Question

[Kathy Nguyễn]

Tìm x

1) \(4x^2+4x+6y+9y^2+2=0\)

2). \(25x^2+9y^2-10x+12y+5=0\)

3). \(9x^2+4y^2+12x-8y+17=0\)

Answer 1

[Mysterious Person]

1) \(4x^2+4x+6y+9y^2+2=0\Leftrightarrow\left(4x^2+4x+1\right)+\left(9y^2+6y+1\right)=0\)

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

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

vậy \(x=\dfrac{-1}{2};y=\dfrac{-1}{3}\)

2) \(25x^2+9y^2-10x+12y+5=0\Leftrightarrow\left(25x^2-10x+1\right)+\left(9y^2+12y+4\right)=0\)

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

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

vậy \(x=\dfrac{1}{5};y=\dfrac{-2}{3}\)

3) \(9x^2+4y^2+12x-8y+17=0\Leftrightarrow\left(9x^2+12x+4\right)+\left(4y^2-8y+4\right)+9=0\)

\(\Leftrightarrow\left(3x+2\right)^2+\left(2y-2\right)^2+9=0\)

ta có : \(\left(3x+2\right)^2\ge0\forall x\) và \(\left(2y-2\right)^2\ge0\forall y\)

\(\Rightarrow\) \(\left(3x+2\right)^2+\left(2y-2\right)^2+9\ge9>0\forall x;y\)

\(\Rightarrow\) phương trình vô nghiệm