\(\hept{\begin{cases}\left|x+\frac{1}{2009}\right|\ge0\\....\\\left|x+\frac{2008}{2009}\right|\ge0\end{cases}\Rightarrow\left|x+\frac{1}{2009}\right|+\left|x+\frac{2}{2009}\right|+....\left|x+\frac{2008}{2009}\right|\ge0}\)

\(\Rightarrow2009x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2009}\right|=x+\frac{1}{2009}\\....\\\left|x+\frac{2008}{2009}\right|=x+\frac{2008}{2009}\end{cases}\Rightarrow x+\frac{1}{2009}+...+x+\frac{2008}{2009}}=2009x\)

\(2008x+201840=2009x\Rightarrow x=201840\)

p/s: cách làm thì khá ok, nhưng kq không chắc lắm nhé, có gì bn tính lại nha

Boul đẹp trai_tán gái đổ 100% sai 100%

Sao dòng cuối lại tek ? Các phân số ấy cộng vào không thể là 201840

Về hướng làm thì đúng nhưng chỉ đúng đến bước phá trị thôi 

Tham khảo cách làm nhưg nhớ đổi đoạn cuối nhé !

a sorry cộng lại quên mẹ chia cho 2009 :> mà tính máy tính ko hiểu sao cx sai lun, đổi lại kq nha :>

\(x=\frac{2017036}{2009}=1004\)

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) \(0,25x^3+x^2+x=0\)

\(\Leftrightarrow x\left(0,25x^2+x+1\right)=0\)

\(\Leftrightarrow x\left[\left(\frac{1}{2}x\right)^2+2\cdot\frac{1}{2}x\cdot1+1^2\right]=0\)

\(\Leftrightarrow x\left(\frac{1}{2}x+1\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

Vậy....

b) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)

\(\Leftrightarrow\frac{2-x}{2007}-1+2=\frac{1-x}{2008}+1+\frac{-x}{2009}+1\)

\(\Leftrightarrow\frac{2-x+2007}{2007}=\frac{1-x+2008}{2008}+\frac{-x+2009}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}-\frac{2009-x}{2008}-\frac{2009-x}{2009}=0\)

\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

\(\Rightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

Vậy....

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)

\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)

\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)

\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)

\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)

x=-2011

cách bạn giải bài này là gì vậy ?

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

\(\Leftrightarrow\frac{x-2008-2009-2010}{2008}+\frac{x-2008-2009-2010}{2009}+\frac{x-2008-2009-2010}{2010}=0\)

\(\Leftrightarrow\left(x-2008-2009-2010\right)\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)

\(\Leftrightarrow x-6027=0\Leftrightarrow x=6027\)

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

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

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

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

\(=>x+2010=0\)

\(=>x=-2010\)

\(pt\Leftrightarrow\frac{x}{2009}+\frac{1}{2009}+\frac{x}{2008}+\frac{2}{2008}=\frac{x}{3}+\frac{2007}{3}+\frac{x}{4}+\frac{2006}{4}\Leftrightarrow\frac{x}{2009}+\frac{x}{2008}-\frac{x}{3}-\frac{x}{4}=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\right)=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x=\frac{\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}}{\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}}=-2010\)

1) =0

3)=(3⁷⁹⁸⁸-1)/2

3) Q=(3+1)(3^2+1)(3^4+1)....(3^3994+1)

=(3-1)(3+1)(3^2+1)(3^4+1)...(3^3994+1)

=(3^2-1)(3^2+1)(3^4+1)...(3^3994+1)

=(3^4-1)(3^4+1)...(3^3994+1)

=.........

=(3^3994-1)(3^3994+1)

=3^7988-1

quá khó vì em mới học lớp 6