Tìm x, y, z
a) \(2009-\left|x-2009\right|=x \)
b) \(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x-+y-z\right|=0\)
Tìm x; y; z :
a) \(2009-\left|x-2009\right|=x\)
b) \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
a: =>|x-2009|=2009-x
=>x-2009<=0
=>x<=2009
b: =>2x-1=0 và y-2/5=0 và x+y-z=0
=>x=1/2 và y=2/5 và z=x+y=1/2+2/5=5/10+4/10=9/10
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+\left|x=\frac{1}{20}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
2. Tìm x, y, z biết\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
3.Tìm x\(a,2009-\left|x-2009\right|=x\)
\(b,\left|3x+2\right|=\left|5x-3\right|\)
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
Bài 3:
a)\(2009-\left|x-2009\right|=x\)
\(\Rightarrow\left|x-2009\right|=2009-x\)
\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)
Vì GTTĐ của số âm bằng số đối của nó
\(\Rightarrow x-2009\le0\)
\(\Rightarrow x\le2009\)
Vậy với mọi \(x\le2009\) đều thỏa mãn
b)\(\left|3x+2\right|=\left|5x-3\right|\)
\(\Rightarrow3x+2=5x-3\) hoặc \(3x+2=3-5x\)
\(\Rightarrow2x=5\) hoặc \(8x=1\)
\(\Rightarrow x=\frac{5}{2}\) hoặc \(x=\frac{1}{8}\)
Tìm x biết
a.\(2009-\left|x-2009\right|=x\)
b.\(\left(2x+1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
a)
\(2009-\left|x-2009\right|=x\)
\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)
\(\Rightarrow x-2009\le0\)
\(\Rightarrow x\le2009\)
Vậy \(x\le2009\)
b)
Vì \(\left(2x+1\right)^{2008}\ge0\forall x\)
\(\left(y-\dfrac{2}{5}\right)^{2008}\ge0\forall y\)
\(\left|x+y-z\right|\ge0\forall x,y,z\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\forall x,y,z\)
Mà theo đề bài :
\(\left(2x+1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
\(\Rightarrow\left(2x+1\right)^{2008}=0;\left(y-\dfrac{2}{5}\right)^{2008}=0;\left|x+y-z\right|=0\)
*) Với \(\left(2x+1\right)^{2008}=0\)
\(\Rightarrow2x+1=0\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\)
*) Với \(\left(y-\dfrac{2}{5}\right)^{2008}=0\)
\(\Rightarrow y-\dfrac{2}{5}=0\)
\(\Rightarrow y=\dfrac{2}{5}\)
*) Với \(\left|x+y-z\right|=0\)
\(\Rightarrow x+y-z=0\)
\(\Rightarrow\dfrac{-1}{2}+\dfrac{2}{5}-z=0\)
\(\Rightarrow\dfrac{-1}{10}-z=0\)
\(\Rightarrow z=\dfrac{-1}{10}\)
Vậy \(x=\dfrac{-1}{2};y=\dfrac{2}{5};z=\dfrac{-1}{10}\)
a, 2009 - \(\left|x-2009\right|\) = x
=> \(\left|x-2009\right|\) = 2009 - x
=> \(\left[{}\begin{matrix}x-2009=2009-x\\x-2009=-2009-x\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4018\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2009\\x=0\end{matrix}\right.\)
Vậy x \(\in\)n { 2009 ; 0 }
Câu 1:Tìm x:
a)\(2009-\left|x-2009\right|=x\)
b)\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
Câu 2: Tìm 3 số a;b;c biết:
\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\) và a + b + c = -50
Ta thấy \(\left\{\begin{matrix}\left(2x-1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\\\left|x+y-z\right|\ge0\end{matrix}\right.\ge0\)
\(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\)
Mà theo đề ra
\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
\(\Rightarrow\left\{\begin{matrix}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}2x=1\\y=\frac{2}{5}\\z=x+y\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{2}{5}+\frac{1}{2}=\frac{9}{10}\end{matrix}\right.\)
Vậy \(x=\frac{1}{2}\) y=\(\frac{2}{5}\)và z=\(\frac{9}{10}\)
a)\(2009-\left|x-2009\right|=x\)
\(\Rightarrow\left|x-2009\right|=2009-x\)
\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)
\(\Rightarrow x-2009\le0\)
\(\Rightarrow x\le2008\)
Tìm x,y,z biết
a,\(2009-|x-2009|=x\)
b,\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+|x+y-z|=0\)
---> Các bạn giúp mk nha<---
a) \(2009-\left|x-2009\right|=x\)
* Nếu \(x-2009\ge0\Rightarrow x\ge2009\)
\(2009-\left(x-2009\right)=x\)
\(2009-x+2009=x\)
\(4018=2x\)
\(x=2009\)(TMĐK)
* Nếu \(x-2009< 0\Rightarrow x< 2009\)
\(2009-\left[-\left(x-2009\right)\right]=x\)
\(2009-\left(-x+2009\right)=x\)
\(2009+x-2009=x\)
\(0x=0\)
Nên \(x\in R\) trừ \(x< 2009\)
Vậy .......
1. Tìm x ; y ; z biết :
a) \(2009-\left|x-2009\right|=x\)
b) \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
2. Tìm a,b,c biết :
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\) và a+b+c=-50
1. a) \(2009-\left|x-2009\right|=x\)
\(\Rightarrow\left|x-2009\right|=2009-x\)
\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)
\(\Rightarrow x-2009\le0\)
\(\Rightarrow x\le2009\)
Vậy \(x\le2009.\)
b) Ta có: \(\left[{}\begin{matrix}\left(2x-1\right)^{2008}\ge0\forall x\\\left(y-\dfrac{2}{5}\right)^{2008}\ge0\forall y\\\left|x+y-z\right|\ge0\forall x,y,z\end{matrix}\right.\) \(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\forall x,y,z\)
Dấu \("="\) xảy ra khi \(\left[{}\begin{matrix}\left(2x-1\right)^{2008}=0\\\left(y-\dfrac{2}{5}\right)^{2008}=0\\\left|x+y-z\right|=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\).
Bạn kia làm câu 1 rồi thì mình làm câu 2 nhé!
2. Ta có:\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\)
\(\Rightarrow\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{5b-3c}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{15a-10b+6c-15a}{25+9}\)=\(\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)
\(\Rightarrow\dfrac{-5b+3c}{17}=\dfrac{5b-3c}{2}\Rightarrow5b-3c=0\)
=> 5b=3c =>\(\left\{{}\begin{matrix}b=\dfrac{3}{5}c\\a=\dfrac{2}{5}c\end{matrix}\right.\)
=>\(\dfrac{3}{5}c+\dfrac{2}{5}c+c=-50\)
=> \(c\left(\dfrac{3}{5}+\dfrac{2}{5}+1\right)=-50\)
=> 2c = -50
=> c= -25
=>\(\left\{{}\begin{matrix}b=-25.\dfrac{3}{5}=-15\\a=-25.\dfrac{2}{5}=-10\end{matrix}\right.\)
Vậy a= -10; b= -15; c= -25
\(\left|x+\frac{13}{7}\right|+\left|y+\frac{2009}{2008}\right|+\left|z+2007\right|=0\)
Tìm x, y, z
Vì giá trị tuyệt đối của từng cái đó luôn lớn hơn hoặc bằng không nên biểu thức đó bằng không khi:
\(\left|x+\frac{13}{7}\right|=0\Rightarrow x+\frac{13}{7}=0\Rightarrow x=\frac{-13}{7}\)
\(\left|y+\frac{2009}{2008}\right|=0\Rightarrow y+\frac{2009}{2008}=0\Rightarrow y=\frac{-2009}{2008}\)
\(\left|z+2007\right|=0\Rightarrow z+2007=0\Rightarrow z=-2007\)
Vậy ....
Tìm x y z biết
\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
(2x - 1 )2008+(y - 2/5)2008 + |x + y - z | = 0
=> ( 2x - 1) 2008 =0 => 2x - 1 =0 => 2x = 1 => x = 1/2
( y - 2/5 )2008 = 0 y - 2/5 = 0 y =2/5 y = 2/5
|x + y -z | = 0 x + y - z = 0 x + 2/5 - z = 0 1/2 - 2/5 -z = 0
=>x = 1/2 =>x = 1/2
y = 2/5 y = 2/5
5/10 - 4/10 = z z = 1/ 10
Vậy x = 1/2 ; y = 2/5 : z = 1/10
( nhớ cho mk nha )
ta có: \(\left(2x-1\right)^{2008}\ge0\)
\(\left(y-\frac{2}{5}\right)^{2008}\ge0\)
\(\left|x+y-z\right|\ge0\)
\(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\)
để \(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
\(\Rightarrow\left(2x-1\right)^{2008}=0\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)
\(\left(y-\frac{2}{5}\right)^{2008}=0\Rightarrow y-\frac{2}{5}=0\Rightarrow\frac{2}{5}\)
\(\left|x+y-z\right|=0\Rightarrow x+y-z=0\Rightarrow z=x+y\Rightarrow z=\frac{1}{2}+\frac{2}{5}=\frac{9}{10}\)
KL: x= 1/2; y= 2/5; z=9/10
( mk nghĩ nó còn có nhiều đáp số lắm, nhưng mk ko bít cách lm)
Do (2x-1)2008\(\ge0\),\(\left(y-\frac{2}{5}\right)^{2008}\ge0\),|x+y-z|\(\ge0\)
mà đề cho tổng 3 số trên bằng 0
\(\Rightarrow\hept{\begin{cases}\left(2x-1\right)^{2008}\\\left(y-\frac{2}{5}\right)^{2008}\\\left|x+y-z\right|=0\end{cases}\Rightarrow\hept{\begin{cases}2x=-1\\y=\frac{2}{5}\\x+y-z=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\\frac{1}{2}+\frac{2}{5}-z=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)
Vậy ...(bn tự kl nhé)
1.TÌm x,y,z biết
a.2009-\(\left|x-2009\right|\)=x
b.\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+u-z\right|\)=0
2.Tìm các số a,b,c biết
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\)
và a+b+c = -50
Ta luôn có :|x-2009|\(\ge\)0(1)
Mà :2009-|x-2009|=x nên 2009\(\ge\)x(2)
Vì (1)và(2) nên ta có x \(\in\){0;1;2;3;4;5;...;2009}