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

Tag thầy Lâm không :)???

\(\Rightarrow\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x=-3\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\ne0\right)\)

\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)

\(\Leftrightarrow x+3=0\)

hay x=-3

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

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

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

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}

Áp dụng bđt Cauchy Shwarz dạng Engel, ta có:

\(\dfrac{x^4}{a}+\dfrac{y^4}{b}\ge\dfrac{\left(x^2+y^2\right)^2}{a+b}=\dfrac{1}{a+b}\) (vì \(x^2+y^2=1\))

mà \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}\) (theo đề bài)

\(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\) (tính chất của dãy tỉ số bằng nhau)

\(\Rightarrow x^2=\dfrac{a}{a+b}\)

\(B=\dfrac{x^{2008}}{a^{1004}}+\dfrac{y^{2008}}{b^{1004}}\)

\(=\left(\dfrac{x^2}{a}\right)^{1004}+\left(\dfrac{y^2}{b}\right)^{1004}\)

\(=2\times\left(\dfrac{\dfrac{a}{a+b}}{a}\right)^{1004}\) (vì \(\dfrac{x^2}{a}=\dfrac{y^2}{b}\))

Thay số vào ròi tính thoy ~~! (xxx)

\(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{\left(x^2+y^2\right)^2}{a+b}\left(dox^2+y^2=1\right)\)

\(\Leftrightarrow\dfrac{x^4}{a}+\dfrac{y^4}{b}-\dfrac{\left(x^2+y^2\right)^2}{a+b}=0\)

Tự biến hóa, hô phép ;v

\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\)

\(\Rightarrow\dfrac{x^{2008}}{a^{1004}}=\dfrac{y^{2008}}{b^{2004}}\Rightarrow B=2.\dfrac{x^{2008}}{a^{1004}}\)

\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)

\(\Rightarrow\dfrac{x^{2008}}{a^{1004}}=\dfrac{1}{\left(a+b\right)^{1004}}\)

\(\dfrac{2}{\left(a+b\right)^{1004}}=2.\dfrac{x^{2008}}{a^{1004}}=B\)

Vậy: \(B=\dfrac{2}{\left(a+b\right)^{1004}}\)

a: \(A=\dfrac{6}{7}x^2y^2\cdot\dfrac{-7}{2}x^2y=-3x^4y^3\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:

\(\dfrac{x}{-2}=\dfrac{y}{3}=\dfrac{x-y}{-2-3}=\dfrac{5}{-5}=-1\)

Do đó: x=2; y=-3

\(A=-3x^4y^3=-3\cdot2^4\cdot\left(-3\right)^3=3\cdot27\cdot16=81\cdot16=1296\)

\(A=\dfrac{6}{7}x^2y^2.\left(-3\dfrac{1}{2}x^2y\right)\)

\(=\dfrac{6}{7}x^2y^2.\left(-\dfrac{7}{2}\right)x^2y\)

\(=-3x^4y^3\)

b)Có: \(\dfrac{x}{y}=-\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{-y}{3}=\dfrac{x-y}{2+3}=\dfrac{5}{5}=1\)

\(\Rightarrow x=2;y=-3\)

Tại \(x=2;y=-3\) , giá trị của biểu thức là:

 \(-3.2^4.\left(-3\right)^3=-3.16.\left(-27\right)=1296\)