Tìm x,y,z biết:
a) 2019 - |x -2019| = x
b) \(^{\left(2x-1\right)^{2008}}\) + (y - \(\dfrac{2}{5}\))^2008 + |x + y - z| = 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
tìm x,y,z biết:a)2019-|x-2019|=x
b)(2x-1)2008+(y-2/5)2008+|x+y+z|=0
ta có: |x-2019|>=0 với mọi x
=>2019-|x-2019|<=2019-0=2019 với mọi x
=>x<=2019
=>2019-|x-2019|=2019-2019-x=-x=x
=>x=0
a. Tím số x,y,z biết: \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+|x+y-z|=0\)
b. Tìm x sao cho: \(\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}^{x+4}\right)=17\)
Nhờ anh em tiếp nha.
a: \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
nên \(\left\{{}\begin{matrix}2x-1=0\\y-\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=x+y=\dfrac{9}{10}\end{matrix}\right.\)
b: Bạn xem lại đề, nghiệm rất xấu
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)
Tìm x,y,z biết
\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
Giúp mk vs!!!
\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
Nhận xét : \(\left\{{}\begin{matrix}\left(2x-1\right)^{2008}\ge0\forall x\\\left(y-\dfrac{2}{5}\right)^{2008}\ge0\forall y\\\left|x+y+z\right|\ge0\forall x,y,z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x-1\right)^{2008}=0\\\left(y-\dfrac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=-\dfrac{9}{10}\end{matrix}\right.\)
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 }
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
Tìm x,y,z biết:
a) \(2019-|x-2019|=x\)
b) \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}+|x+y-z|=0\)
a)\(2019-\left|x-2019\right|=x\)
\(\Rightarrow2019-x=\left|x-2019\right|\)
=>\(\left|x-2019\right|=-\left(x-2019\right)\)
=>\(x-2019\le0\)
=>\(x\le2019\)
b) Vì \(\left(2x-1\right)^{2018}\ge0\forall x\)
\(\left(y-\frac{2}{5}\right)^{2018}\ge0\forall y\)
\(\left|x+y-z\right|\ge0\forall x,y,z\)
=> \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|\ge0\forall x,y,z\)
mà \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}}\)=>\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}\)
a, Ta có:
\(\left|x-2019\right|=\orbr{\begin{cases}x-2019\ge0\Rightarrow x\ge2019\\-x+2019< 0\Rightarrow x< 2019\end{cases}}\)
Xét x<2019 thì |x-2019|=-x+2019
Khi đó: 2019-(-x+2019)=x
\(\Leftrightarrow\)-x+2019=2019-x
\(\Leftrightarrow\)-x+2019+x=2019
\(\Leftrightarrow\)0x+2019=2019
\(\Leftrightarrow\)0x=0 (thỏa mãn)
Xét 2019\(\le\)x thì |x-2019|=x-2019
Khi đó 2019-(x-2019)=x
\(\Leftrightarrow\)2019-x+2019=x
\(\Leftrightarrow\)4038-x=x
\(\Leftrightarrow\)4038=2x
\(\Leftrightarrow\)x=2019(thỏa mãn)
Vậy .......................................................!!!
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 .......