Question

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

Answer 1

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 \(x=\dfrac{1}{4};y=-\dfrac{1}{4}\)

Answer 2

Nguyễn Huy TúAce LegonaNghiêm Gia Phương

Hoàng Thị Ngọc AnhHoang Hung QuanngonhuminhĐức Huy ABC

An NguyễnNguyễn Nhật MinhPhạm Nguyễn Tất Đạt

Đức Minh

Answer 3

GTNN là j thế???