\(\left|2\text{x}-3\right|-x=\left|2-x\right|\)
=> \(2\text{x}-3-x=2-x\) hoặc \(2\text{x}-3-x=-2+x\)
TH1 :
\(2\text{x}-3-x=2-x\)
=> 2x - 3 - x - 2 + x=0
=> 2x - 3 -2 = 0
=> 2x - 5 =0
=> 2x = 5
=> x = \(\dfrac{5}{2}\)
TH2 :
\(2\text{x}-3-x=-2+x\)
=> \(2\text{x}-3-x+2-x\) = 0
=> 0x -3 + 2 = 0
=> 0x - 1 = 0
=> 0x = 1 ( Không có x )
Vậy : x = \(\dfrac{5}{2}\)
Với giá trị nào của x mỗi biểu thức sau đạt giá trị nhỏ nhất:
\(a)A=\left|x-3\right|+\left|x-5\right|+\left|x-7\right|\)
\(b)B=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x+5\right|\)
\(c)C=\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|+\left|x+5\right|\)
\(d)D=\left|2x+1\right|+\left|2x+3\right|+\left|2x+5\right|+\left|2x+7\right|+\left|2x+9\right|+\left|2x+11\right|\)
***** ___ GIÚP MK VS MN ƠI___ MK CẦN GẤP___ *****
\(a.\left|x+2\right|+\left|x-1\right|=3-\left(y+2\right)^2\)
\(b.\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
\(c.\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
\(d.\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y-3\right|+3}\)
Bài 4 : Tìm x,y biết :
a)\(3\dfrac{1}{4}-\dfrac{1}{4}\left|x-0,25\right|=\dfrac{1}{4}\)
b)\(\left|x-3\right|=-4\)
c)\(\left|x-\dfrac{1}{2}\right|+1=0\)
d)\(\left|x-\dfrac{1}{5}\right|=\left|\dfrac{3}{4}-2x\right|\)
e)\(\left|x+2\right|3=0\)
f)\(\left|2x-1\right|-\left|x+2\right|=0\)
g)\(\left|x+2\right|+\left|y-1\right|=0\)
h)\(\left|2x\right|+\left|y+\dfrac{1}{4}\right|=0\)
Bài 1,Tìm x,biết:
a, \(4\left|3x-1\right|+\left|x\right|-2\left|x-5\right|+7\left|x-3\right|=12\)
b, \(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
Tính giá trị của biểu thức
a) \(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-\left|x\right|\right)+y\) với x = 3 và y = -2
b) \(B=\left|2x-1\right|+\left|3y+2\right|\) với x = 3 và y = -3
1, \(\left|2x-3\right|\)= \(\left|x+5\right|\)
2, \(\left|4-2x\right|\)=\(\left|3x\right|\)
3, \(\left|4x-5\right|\) -\(\left|2x+1\right|\) =0
4, \(\left|0,5x-2\right|\) -\(\left|x+\frac{2}{3}\right|\) =0
Tìm GTNN của các biểu thức sau:
F=\(\left|2x-2\right|+\left|2x-2003\right|\)
G=\(\left|2x-3\right|+\frac{1}{2}\left|4x-1\right|\)
H=\(\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\)
CÁC BẠN GIÚP MÌNH VỚI !!!!!!!!!!!!!!!!!
1/ Tìm GTNN:
a,\(\left|7x-5y\right|+\left|2z-3y\right|+\left|xy+yz+xz-2000\right|\)
b,\(\left|3x-7\right|+\left|3x+2\right|+8\)
2/ Tìm x, y:
a, \(\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
b,\(\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
c,\(\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}\)
Tìm x, biết:
1) \(\left|2x-3\right|=2x-3\)
2) \(\left|5x-\dfrac{2}{3}\right|=\dfrac{2}{3}-5x\)
3) \(\left|3-x\right|+\left|2y-5\right|\le0\)
Tìm x biết :
a )\(\left|2x-1\right|=1,5\)
b)\(7,5-\left|5-2x\right|=-4,5\)
c)\(-3+\left|x\right|=-1\)
d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)
e)\(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)
f)\(\left|2x+3\dfrac{1}{5}\right|-\dfrac{1}{2}=\dfrac{3}{10}\)
