Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)
\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+...+\dfrac{x+50}{50}+1=0\)
\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}\right)=0\)
mà \(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}>0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+...+\left(\dfrac{x+50}{50}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}\right)=0\)
\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}>0\) )
\(\Leftrightarrow x=-100\)
