\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)
\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)
\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=0\)
\(\Leftrightarrow x-2000=0\).Do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\ne0\)
\(\Leftrightarrow x=2000\)
\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)
\(\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+3=3\)
\(\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)
\(\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=0\)
Mà: \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\ne0\)
\(\Rightarrow\)x-2000=0
Vậy : x=2000
