ĐK : 2008 \leq x \leq 2010
2010−x+x−2008=x2−4018x+4036083
2010−x−1+x−2008−1−(x2−4018x+4036081)=0
x−2008+1x−2009−2010−x+1x−2009−(x−2009)2=0
(x−2009)(x−2008+11−2010−x+11+2009−x)=0
\Leftrightarrow x= 2009
Hoặc
x−2008+11−2010−x+11+2009−x=0
pt trên có nghiệm duy nhất x = 2009
Vậy pt đã cho có nghiệm x = 2009
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