\(\frac{x+2}{2008}\)+ 1 + \(\frac{x+3}{2007}\)+1 +\(\frac{x+4}{2006}\)+1 +\(\frac{x+2028}{6}\)-3=0
\(\Leftrightarrow\)\(\frac{x+2+2008}{2008}\)+ \(\frac{x+3+2007}{2007}\) + \(\frac{x+4+2006}{2006}\)+ \(\frac{x+2028-18}{6}\)= 0
\(\Leftrightarrow\) \(\frac{x+2010}{2008}\)+ \(\frac{x+2010}{2007}\)+ \(\frac{x+2010}{2006}\)+ \(\frac{x+2010}{6}\)=0
\(\Leftrightarrow\)(x +2010).\(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)\)=0
\(\Leftrightarrow\)x + 2010 = 0 \(\left(vì\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\right)\)
\(\Leftrightarrow\) x = -2010
Vậy S = \(\left\{-2010\right\}\)
⇔ \(\dfrac{x+2}{2008}\) +1 +\(\dfrac{x+3}{2007}\) +1+\(\dfrac{x+4}{2006}\)+1 +\(\dfrac{2028}{6}\)-3 =0
⇔\(\dfrac{x+2}{2008}+\dfrac{2008}{2008}+\dfrac{x+3}{2007}+\dfrac{2007}{2007}+\dfrac{x+4}{2006}+\dfrac{2006}{2006}+\dfrac{x+2028}{6}-\dfrac{18}{6}=0\)
⇔\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
⇔(x+2010)\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)
Mà \(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)\)≠0
⇒x+2010=0
⇔x=-2010
Vậy phương trình có nghiệm x=-2010
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Câu hỏi của Silver Bullet - Toán lớp 8 | Học trực tuyến