Question

[Đinh Thị Thùy Trang]

Tìm x,y biết:\(\left(2x-\frac{1}{6}\right)^2+\left|3y+12\right|\le0\)

Answer 1

[Edogawa Conan]

Ta có: \(\left(2x-\frac{1}{6}\right)^2\ge0\forall x\)

        \(\left|3y+12\right|\ge0\forall y\)

=> \(\left(2x-\frac{1}{6}\right)^2+\left|3y+12\right|\ge0\forall x;y\)

=> \(\hept{\begin{cases}2x-\frac{1}{6}=0\\3y+12=0\end{cases}}\)

=> \(\hept{\begin{cases}2x=\frac{1}{6}\\3y=-12\end{cases}}\)

=> \(\hept{\begin{cases}x=\frac{1}{12}\\y=-4\end{cases}}\)