1.TÌm x,y,z biết
a.2009-|x−2009||x−2009|=x
b.\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
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
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
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}
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 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. 8x=5y,7y=12z và x+y+z= -318
b. \(\dfrac{5z-6y}{4}=\dfrac{6x-4z}{5}=\dfrac{4y-5x}{6}\)và 3x-2y+5z=96
c. 2009-|x-2009|=x
d. \(\left(2x-1\right)^{2008}\)+\(\left(y-\dfrac{2^{ }}{5}\right)^{2008}\)+|x+y-z|=0
Vi 8x = 5y , 7y = 12z
=>\(\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{8}\\\dfrac{y}{12}=\dfrac{z}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{60}=\dfrac{y}{96}\\\dfrac{y}{96}=\dfrac{z}{56}\end{matrix}\right.\)
=> \(\dfrac{x}{60}=\dfrac{y}{96}=\dfrac{z}{56}\)
Ap dung tinh chat day ti so bang nhau co
\(\dfrac{x}{60}=\dfrac{y}{96}=\dfrac{z}{56}=\dfrac{x+y+z}{60+96+56}=\dfrac{-318}{212}=\dfrac{-3}{2}\)
\(\dfrac{x}{60}=\dfrac{-3}{2}\Rightarrow x=60.\dfrac{-3}{2}=-90\)
\(\dfrac{y}{96}=\dfrac{-3}{2}\Rightarrow y=96.\dfrac{-3}{2}=-144\)
\(\dfrac{z}{56}=\dfrac{-3}{2}\Rightarrow z=56.\dfrac{-3}{2}=-84\)
Vay x= -90, y= -144 va z=-84
c: =>|x-2009|=2009-x
=>x-2009<=0
=>x<=2009
d: =>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=9/10
a: 8x=5y; 7y=12z
=>x/5=y/8; y/12=z/7
=>x/15=y/24=z/14
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{15}=\dfrac{y}{24}=\dfrac{z}{14}=\dfrac{x+y+z}{15+24+14}=-\dfrac{318}{53}=-6\)
=>x=-90; y=-144; z=-84
Tìm x, y, z
a) \(2009-\left|x-2009\right|=x \)
b) \(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x-+y-z\right|=0\)
a)/x-2009/=2009-x
TH1:x-2009=2009-x=>x=2009
TH2:x-2009=-(2009-x)=>x-2009=x-2009 đúng với mọi x
b) (2x-1)^2008>=0
(y-2/5)^2008>=0
/x-y-z/>=0
=>2x-1=0
y-2/5=0
x-y-z=0(cái này dùng ngoặc nhọn)
=>x=1/2;y=2/5;z=1/10
\(a)\) \(2009-\left|x-2009\right|=x\)
\(\Leftrightarrow\)\(\left|x-2009\right|=2009-x\)
Ta có : \(\left|x-2009\right|\ge0\)
\(\Rightarrow\)\(2009-x\ge0\)
\(\Rightarrow\)\(x\le2009\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2009=2009-x\\x-2009=x-2009\end{cases}\Leftrightarrow\orbr{\begin{cases}x+x=2009+2009\\x=x\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x=4018\\x=x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2009\\x=x\end{cases}}}\)
Vậy \(x=2009\)
Chúc bạn học tốt ~
\(b)\) \(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(2x-1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}2x=1\\y=\frac{2}{5}\\z=x+y\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{2}+\frac{2}{5}\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}\)
Vậy nghiệm của phương trình là \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}\)
Chúc bạn học tốt ~
Tìm giá trị lớn nhất , giá trị nhỏ nhất của biểu thức :
a)\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\)
b)B=\(\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\)
c)C=\(-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
Ai lm đc câu nào thì giúp mk với , cảm ơn !!
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)