Question

Tìm x và y biết: 6x² + 18y² +12x -12xy +9=0

Answer 1

\(6x^2+18y^2+12x-12xy+9=0\)

\(\Rightarrow\left(2x^2-12xy+18y^2\right)+\left(4x^2+12x+9\right)=0\)

\(\Rightarrow2\left(x^2-6xy+9y^2\right)+\left(2x+3\right)^2=0\)

\(\Rightarrow2\left(x-3y\right)^2+\left(2x+3\right)^2=0\)

Vì \(\left\{{}\begin{matrix}2\left(x-3y\right)^2\ge0\\\left(2x+3\right)^2\ge0\end{matrix}\right.\) với mọi x,y

=> \(2\left(x-3y\right)^2+\left(2x+3\right)^2\ge0\)

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}3y=x\\2x=-3\end{matrix}\right.\)

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

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