\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
\(\Rightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)
\(\Rightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=0\)
Mà \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\ne0\)
\(\Rightarrow x-2000=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000
Có : \(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
\(\Leftrightarrow\)\(\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)
\(\Leftrightarrow\) \(\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)
\(\Leftrightarrow\) \(\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=0\)
\(\Leftrightarrow\) \(x-2000=0\)
( vì \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)\ne0\) )
\(\Leftrightarrow\) \(x=2000\)
Vậy x = 2000 thì phương trình \(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
