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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
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}
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 sai đề
b) Ta có:
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\Leftrightarrow\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}\)Hay \(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{10b-6c}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{10b-6c}{4}=\dfrac{15a-10b+6c-15a+10b-6c}{25+9+4}=\dfrac{0}{25+9+4}=0\)
Nên
\(\left\{{}\begin{matrix}3a=2b\\2c=5a\\5b=3c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{3}\\\dfrac{c}{5}=\dfrac{a}{2}\\\dfrac{b}{3}=\dfrac{c}{5}\end{matrix}\right.\Leftrightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=\dfrac{a+b+c}{2+3+5}=\dfrac{-50}{10}=-5\)
\(\Rightarrow\left\{{}\begin{matrix}a=-5.2=-10\\b=-5.3=-15\\c=-5.5=-25\end{matrix}\right.\)
Câu 3:
Giải:
Ta có: \(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)
\(\Rightarrow\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}=\frac{15a-10b+6c-15a+10-6c}{25+9+4}=\frac{0}{25+9+4}=0\)
\(\Rightarrow\left\{\begin{matrix}\frac{15a-10b}{25}=0\\\frac{6c-15a}{9}=0\\\frac{10b-6c}{4}=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}15a-10b=0\\6c-15a=0\\10b-6c=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}15a=10b\\6c=15a\\10b=6c\end{matrix}\right.\)
\(\Rightarrow15a=10b=6c\)
\(\Rightarrow\frac{15a}{30}=\frac{10b}{30}=\frac{6c}{30}\)
\(\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{2+3+5}=\frac{-50}{10}=-5\)
\(\Rightarrow\left\{\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)
Vậy \(a=-10;b=-15;c=-25\)
3.Từ \(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\)
\(\Rightarrow\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{10b-6c}{4}\)
Áp dụng tc dãy tỉ số bằng nhau ta có :
\(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{10b-6c}{4}=\dfrac{15a-10b+6c-15a+10b-6c}{25+4+9}=\dfrac{0}{25+4+9}=0\)
\(\Rightarrow\left\{{}\begin{matrix}15a-10b=0\\6c-15a=0\\10b-6c=0\end{matrix}\right.\Rightarrow15a=10b=6c\) \(\Rightarrow\dfrac{15a}{60}=\dfrac{10b}{60}=\dfrac{6c}{60}\)
\(\Rightarrow\dfrac{a}{4}=\dfrac{b}{6}=\dfrac{c}{10}\)
Áp dụng tc dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{4}=\dfrac{b}{6}=\dfrac{c}{10}=\dfrac{a+b+c}{4+6+10}=\dfrac{-50}{20}=\dfrac{-5}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)
a, 2009; 0
b, x= 0.5 ; y= 0.4; z=0.9
sai thì thôi nhé
ai lam guip toi cau nay voi mai toi nop bai roi
so sanh 2 phan so sau bang cach nahnh nhat: 2007/2008 voi 2008/2009
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
