Ta có: \(\dfrac{x+1}{200}+\dfrac{x+2}{199}+\dfrac{x+3}{198}=\dfrac{x+200}{1}+\dfrac{x+199}{2}+\dfrac{x+198}{3}\)
=>\(\left(\dfrac{x+1}{200}+1\right)+\left(\dfrac{x+2}{199}+1\right)+\left(\dfrac{x+3}{198}+1\right)=\left(\dfrac{x+200}{1}+1\right)+\left(\dfrac{x+199}{2}+1\right)+\left(\dfrac{x+198}{3}+1\right)\)
=>\(\dfrac{x+201}{200}+\dfrac{x+201}{199}+\dfrac{x+201}{198}=\dfrac{x+201}{1}+\dfrac{x+201}{2}+\dfrac{x+201}{3}\)
=>\(\left(x+201\right)\left(\dfrac{1}{200}+\dfrac{1}{199}+\dfrac{1}{198}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
=>x+201=0
=>x=-201
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