Question

Tìm x , y biết:

\(\left(2x-\dfrac{1}{6}\right)^2+|3y+12|\le0\)

Answer 1

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 !!!