\(\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1=\dfrac{205-x}{95}+1\)
\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}=\dfrac{300-x}{95}\)
\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}-\dfrac{1}{95}\ne0\right)=0\Leftrightarrow x=300\)