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 - \(\left|2x-\dfrac{1}{1009}\right|\)
Bài 3: A=2018-|x+2019|. Vì |x+2019|\(\ge\)0 nên -|x+2019|\(\le\)0=>2018-|x+2019|\(\le\) 2. Vậy A có GTLN = 2 khi x+2019=0 hay x=-2019. B=-10-\(\left|2x-\dfrac{1}{1009}\right|\). Vì \(\left|2x-\dfrac{1}{1009}\right|\ge0\Rightarrow-\left|2x-\dfrac{1}{1009}\right|\le0\Rightarrow-10-\left|2x-\dfrac{1}{1009}\right|\le-10\). Vậy B có GTLN = -10 khi 2x-\(\dfrac{1}{1009}=0\) => \(2x=\dfrac{1}{1009}\Rightarrow x=\dfrac{1}{1009}:2=\dfrac{1}{2018}\)
Bài 2: A=\(\left|5x+1\right|-\dfrac{3}{8}\). Vì \(\left|5x+1\right|\ge0\Rightarrow\left|5x+1\right|-\dfrac{3}{8}\ge\dfrac{-3}{8}\). Vậy A có GTNN = \(\dfrac{-3}{8}\) khi 5x+1= 0=> 5x= -1=> x = \(\dfrac{-1}{5}\). B=\(\left|2-\dfrac{1}{6}x\right|+0,25\) , vì \(\left|2-\dfrac{1}{6}x\right|\ge0\Rightarrow\left|2-\dfrac{1}{6}x\right|+0,25\ge0,25\) . Vậy B có GTNN = 0,25 khi \(2-\dfrac{1}{6}x=0\Rightarrow\dfrac{x}{6}=2\Rightarrow x=2.6=12\)
