Bài 1 : Tìm GTNN
A = \(\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|+\dfrac{13}{14}\)
Bài 2 : Tìm x , y là số nguyên
a) \(\dfrac{x+3}{5}-\dfrac{1}{y-1}=\dfrac{2}{5}\)
b) \(-1-\dfrac{7}{14}\le\dfrac{x}{3}\le1+\dfrac{12}{48}\)
c) \(\dfrac{2y+7}{y-2}\in Z\)
giúp mk với 
sáng mai mk có tiết toán
Bài 1:
Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2\ge0\\\left|y+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\Rightarrow\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|\ge0\)
\(\Rightarrow A=\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|+\dfrac{13}{14}\ge\dfrac{13}{14}\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2=0\\\left|y+\dfrac{1}{4}\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{4}=0\\y+\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-1}{4}\end{matrix}\right.\)
Vậy \(MIN_A=\dfrac{13}{14}\) khi \(x=\dfrac{1}{4};y=-\dfrac{1}{4}\)
