Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2\ge0\forall x\\\left|y+\dfrac{1}{4}\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|+\dfrac{13}{14}\ge\dfrac{13}{14}\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(Min_A=\dfrac{13}{14}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2\ge0\\\left|y+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\)\(\forall x,y\)
\(\Rightarrow\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|\ge0\forall x,y\)
\(\Rightarrow\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|+\dfrac{13}{14}\ge\dfrac{13}{14}\forall x,y\)
\(\Rightarrow A\ge\dfrac{13}{14}\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2=0\\\left|y+\dfrac{1}{4}\right|=0\end{matrix}\right.\)\(\Rightarrow x=y=\pm\dfrac{1}{4}\)
Vậy với \(x=y=\pm\dfrac{1}{4}\) thì \(A_{Min}=\dfrac{13}{14}\)
sáng mai mk có tiết toán
