Tìm x, y biết \(|x-2007|+|x-2008|-|x-2009|+|x-2010|=3\)
Tìm x, y biết |x-2007|+|x-2008|+|y-2009|+|x-2010|=3
!x-2007!+!x-2010!>=3 đẳng thức khi 2007<=x<=2008
!x-2007!+!x-2008!+!x-2010!>=3 đẳng thức khi !x-2008!=0
=> nghiệm duy nhất x=2008 và y=2009
Tìm x,y biết
\(|x-2007|+|x-2008|+|y-2009|+|x-2010|=3\)
\(\left|x-2007\right|+\left|x-2010\right|+\left|x-2008\right|+\left|y-2009\right|\)
\(\ge\left|x-2007+2010-x\right|+\left|x-2008\right|+\left|y-2009\right|=3+0+0=3\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-2007\right)\left(2010-x\right)\ge0\\\left|x-2008\right|=0\\\left|y-2009\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2008\\y=2009\end{cases}}\)
Vậy x = 2008 và y = 2009
\(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\)
\(\Rightarrow\left|x-2017\right|+\left|x-2018\right|+\left|2010-x\right|+\left|y-2009\right|=3\)
Ta có :+) \(\left|x-2007\right|+\left|2010-x\right|\ge\left|x-2007+2010-x\right|=3\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2007\right)\left(2010-x\right)\ge0\Leftrightarrow2007\le x\le2010\)
+) \(\left|x-2008\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow x-2008=0\Leftrightarrow x=2008\)
+)\(\left|y-2009\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow y-2009=0\Leftrightarrow y=2009\)
\(\Rightarrow\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|\ge3\)
\(\Rightarrow\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\)
\(\Leftrightarrow\hept{\begin{cases}2007\le x\le2010\\x=2008\\y=2009\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2008\\y=2009\end{cases}}\)
Vậy................................
|x-2007|+|x-2008|+|y-2009|+|x-2010|=3 Tìm x y
tìm x biết
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
giúp mk nhó
\(\Rightarrow\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x=-3\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\ne0\right)\)
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Tìm x∈Z biết
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2007}=\dfrac{x+2}{2009}+\dfrac{x+1}{2010}\)
Tìm x nguyên để M=/x-3/+/x-5/ đạt giá trị nhỏ nhất
Tìm x,y biêt /x-2007/+/x-2008//y-2009/+/x-2010/=3
Tìm x,biết
x+4/2007 + x+3/2008 = x+2/2009 + x+1/2010
(x-4)/2007 + (x-3)/2008)= (x-2)/2009 + (x-1)/2010
=[(x-4)/2007 -1]+[(x-3)/2008 -1]=[(x-2)/2009 -1]+[(x-1)/2010 -1]
=(x-2011)/2007+(x-2011)/2008=(x-2011)/...
=(x-2011)(1/2007+1/2008-1/2009-1/2010)...
suy ra x=2010
Tìm X biết: \(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}+\frac{x+3}{2009}\)
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}+\frac{x+3}{2009}\)
=> \(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}-\frac{x+3}{2009}=0\)
=> \(\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
=> x + 3 = 0
=> x = 0 - 3
=> x = -3
tìm x:
x+3/2007-x+3/2008=x+3/2010-x+3/2009
giải giúp mình nha
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)
\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\) ( trừ 2 số bằng nhau)
\(\Rightarrow\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)
Mà \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\)
\(\Rightarrow x+3=0\)
\(\Rightarrow x=0-3=-3\)