TH1: x<-1/2
A=2(1009-x)-2x-1=-4x+2017
TH2: -1/2<=x<1009
A=2x+1+2(1009-x)=2019
TH3: x>=1009
A=2(x-1009)+2x+1=4x-2017
TH1: x<-1/2
A=2(1009-x)-2x-1=-4x+2017
TH2: -1/2<=x<1009
A=2x+1+2(1009-x)=2019
TH3: x>=1009
A=2(x-1009)+2x+1=4x-2017
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 - \(\left|2x-\dfrac{1}{1009}\right|\)
\(A=\left(\dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009}\right):\left(\dfrac{1}{23}+\dfrac{1}{7}-\dfrac{1}{1009}+\dfrac{1}{7}.\dfrac{1}{23}.\dfrac{1}{1009}\right)+1:\left(30.1009-160\right)\)
tinh
\(A=\left(\frac{1}{7}+\frac{1}{23}-\frac{1}{1009}\right):\left(\frac{1}{23}+\frac{1}{7}-\frac{1}{1009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{1009}\right)+1:\left(30.1009-160\right)\)
1: \(\left(x-2\right)^2-2\cdot\left(x+1\right)^2=\left(2x+1\right)\cdot\left(1-3x\right)-2x\cdot\left(1-x\right)\)
1: \(15-\left\{2\cdot\left[x-\left(2x-4\right)\cdot5\right]\cdot3\cdot\left(x+1\right)\right\}=12-x\)
1: \(\dfrac{2\cdot\left(x+2\right)}{3}-\dfrac{5\cdot\left(x-1\right)}{4}=\dfrac{3\cdot\left(5-x\right)}{2}-1\dfrac{1}{2}\cdot\left(2x+3\right)\)
Tìm min, max:
a)\(I=2011.\left|2x-4\right|+2012.\left(y+1\right)^2+\left(-1\right)\)
b)\(J=\frac{2}{\left(x+2\right)^2+3}\)
Tìm x trong các đẳng thức:
a) \(\left|2x-3\right|=5\)
b) \(\left|2x-1\right|=\left|2x+3\right|\)
c) \(\left|x-1\right|+3x=1\)
d) \(\left|5x-3\right|-x=7\)
1, \(4x-2\cdot\left\{x-3\left[2\cdot\left(1-2\right)-3\cdot\left(2x-1\right)\right]\right\}=4-2\left(1-x\right)\)
2, gọi O là 1 điểm của đoạn thẳng AB=4. xác định vị trí O để
a) AB + BO nhỏ nhất
b) AB + BO= 2BO
c) AB + BO = 3130