Bài 1: Tìm x,y biết:
a) \(\left|x-\dfrac{2}{3}\right|+\left|y+x\right|=0\) b) \(\left(x-2y\right)^2+\left|x+\dfrac{1}{6}\right|=0\)
c) \(\left|3x+5y\right|+\left|y-2\right|=0\)
Bài 2: Tìm giá trị nhỏ nhất
A= \(\left|5x+1\right|-\dfrac{3}{8}\) B= \(\left|2-\dfrac{1}{6}x\right|+0,25\)
Bài 3: Tìm giá trị lớn nhất
A= 2018 - \(\left|x+2019\right|\) B= -10 - \(\left|2x-\dfrac{1}{1009}\right|\)
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Bài 1:
a) \(\left|x-\dfrac{2}{3}\right|+\left|y+x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-\dfrac{2}{3}\right|=0\\\left|y+x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\y+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{2}{3}\end{matrix}\right.\)
b) \(\left(x-2y\right)^2+\left|x+\dfrac{1}{6}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\\left|x+\dfrac{1}{6}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x+\dfrac{1}{6}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=x\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=-\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{12}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
Bài 3:
Ta có: \(\left|x+2019\right|\ge0\)
\(\Rightarrow2018-\left|x+2019\right|\le2018\)
\(\Rightarrow A\le2018\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x+2019\right|=0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)
Vậy MaxA = 2018 \(\Leftrightarrow x=-2019\)
a) Ta có: \(|x+2019|\ge0\forall x\)
\(\Rightarrow2018-|x+2019|\le2018-0\)
\(\Rightarrow A\le2018\)
\(\Rightarrow GTNN\)của A=2018
b) Ta có: \(|2x-\dfrac{1}{1009}|\ge0\forall x\)
\(\Rightarrow-10-|2x-\dfrac{1}{1009}|\le-10-0\)
\(\Rightarrow B\le-10\)
\(\Rightarrow GTNN\) của B=-10
