Question

[Kathy Nguyễn]

Tìm x

1). \(9x^2+y^2-10y-12x+29=0\)

2). \(4x^2+12y+29-8x+x^2=0\)

3). \(x^2+29+9y^2+8x-12y=0\)

Answer 1

[Mysterious Person]

1) \(9x^2+y^2-10y-12x+29=0\)

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

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

ta có : \(\left(3x-2\right)^2\ge0\forall x\) và \(\left(y-5\right)^2\ge0\forall y\)

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

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

vậy \(x=\dfrac{2}{3};y=5\)

2) câu này đề sai rồi nha

3) \(x^2+29+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)+9=0\)

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

ta có : \(\left(x+4\right)^2\ge0\forall x\) và \(\left(3y-2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x+4\right)^2+\left(3y-2\right)^2+9\ge9>0\forall x;y\)

vậy phương trình vô nghiệm