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....

Ta có: \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)

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

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

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

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

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

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

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

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

=>2009-x=-

=>x=2009

Vậy tập nghiệm của phương trình S=2009

Lê Chí Cường nhầm đoạn cuối rồi kìa

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

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

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

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

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

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

nên phương trình trên \(\Leftrightarrow2009-x=0\Leftrightarrow x=2009\)

1111111111

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

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

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

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

\(=>x+2010=0\)

\(=>x=-2010\)

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

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

\(\Rightarrow x = \dfrac{2007.2008.2009+2009.2007-2008.2007-2.2008.2009}{2009.2007+2008.2007-2008.2009}\)

sai rồi 

2010 nha

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

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

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

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

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

Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)

=>x+2010=0

=>x=-2010

Vậy x = -2010

chuanhieu cho lam