Vì mũ chẵn và GTTĐ luôn lớn hơn hoặc bằng 0

mà ... ( ghi đề bài ra )

\(\Rightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\\\frac{4}{3}x+\frac{5}{2}y=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{-4}{3}\end{cases}}\)

Vậy,.......

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 ....

2023 =))

a. 2006/2005 x 2007/2006 x 2008/2007 x 2009/2008 x 2010/2009'

= 2006 x 2007 x 2008 x 2009 x 2010 / 2005 x 2006 x 2007 x 2008 x 2009

= 2010/2005

= 402/401

\(\left(1+\frac{1}{2005}\right)x\left(1+\frac{1}{2006}\right)x\left(1+\frac{1}{2007}\right)x\left(1+\frac{1}{2008}\right)x\left(1+\frac{1}{2009}\right)\)

\(=\frac{2006}{2005}x\frac{2007}{2006}x\frac{2008}{2007}x\frac{2009}{2008}x\frac{2010}{2009}\)

\(=\frac{2010}{2005}\)

\(=\frac{402}{401}\)

Nguyễn Khánh Linh

a,

\(\left[1+\frac{1}{2005}\right].\left[1+\frac{1}{2006}\right].\left[1+\frac{1}{2007}\right].\left[1+\frac{1}{2008}\right].\left[1+\frac{1}{2009}\right]\)

  \(\Rightarrow\left[\frac{2005}{2005}+\frac{1}{2005}\right]\left[\frac{2006}{2006}+\frac{1}{2006}\right]\left[\frac{2007}{2007}+\frac{1}{2007}\right]\)  \(\left[\frac{2008}{2008}+\frac{1}{2008}\right]\left[\frac{2009}{2009}+\frac{1}{2009}\right]\)                                                                                                                                                               

\(\Rightarrow\frac{2006}{2005}.\frac{2007}{2006}.\frac{2008}{2007}.\frac{2009}{2008}.\frac{2010}{2009}\)

\(\Rightarrow\frac{2010}{2005}=\frac{402}{401}\)

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 \(x=\frac{1}{8}\)

a)  \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)

     \(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)

     \(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)

     \(\Leftrightarrow x+2010=0\) ( vì 1/2003  +  1/2006  --  1/2011  -- 1/2015   \(\ne\)0)

    \(\Leftrightarrow x=-2010\)

câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<

a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)

<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)

<=> x = 2010

\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)\)

Ta thấy : \(\left|x-1\right|\ge0;\left|x-2\right|\ge0;\left|x-3\right|\ge0\)

=> \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge0\)

=> 4 ( x - 4 ) \(\ge0\). Mà 4 > 0 => \(x-4\ge0=>x\ge4\)hay

\(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=4\left(x-4\right)=>x-1+x-2+x-3=4\left(x-4\right)\) => 3x - 6 = 4x - 16

=> -6+16 = 4x - 3x => x = 10

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 ~ 

Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)

           \(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)

          \(=100.\frac{2}{101}=\frac{200}{101}\)