1.
a) \(\left(2x-\dfrac{1}{6}\right)^2+\left|3y+12\right|\le0\)
Nhận xét : Do \(\left(2x-\dfrac{1}{6}\right)^2\ge0\) với \(\forall x\)
Và \(\left|3y+12\right|\ge0\) với \(\forall y\)
Nên \(\left(2x-\dfrac{1}{6}\right)^2+\left|3y+12\right|\le0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{6}\right)^2+\left|3y+12\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-\dfrac{1}{6}\right)^2=0\\\left|3y+12\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\dfrac{1}{6}=0\\3y+12=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{12}\\y=-4\end{matrix}\right.\)
vậy \(x=\dfrac{1}{12};y=-4\)
tik mik nha !!!