Ta có : \(\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}=\dfrac{2x+93}{7}+\dfrac{2x+94}{6}+\dfrac{2x+95}{5}\)
\(\Leftrightarrow\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}-\dfrac{2x+93}{7}-\dfrac{2x+94}{6}-\dfrac{2x+95}{5}=0\)
\(\Leftrightarrow\dfrac{2x+5}{95}+1+\dfrac{2x+6}{94}+1+\dfrac{2x+7}{93}+1-\dfrac{2x+93}{7}-1-\dfrac{2x+94}{6}-1-\dfrac{2x+95}{5}-1=0\)
\(\Leftrightarrow\dfrac{2x+100}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}-\dfrac{2x+100}{7}-\dfrac{2x+100}{6}-\dfrac{2x+100}{5}=0\)
\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\right)=0\)
Thấy : \(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\ne0\)
\(\Rightarrow2x+100=0\)
\(\Leftrightarrow x=-50\)
Vậy ...
Ta có: \(\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}=\dfrac{2x+93}{7}+\dfrac{2x+94}{6}+\dfrac{2x+95}{5}\)
\(\Leftrightarrow\dfrac{2x+5}{95}+1+\dfrac{2x+6}{94}+1+\dfrac{2x+7}{93}+1=\dfrac{2x+93}{7}+1+\dfrac{2x+94}{6}+1+\dfrac{2x+95}{5}+1\)
\(\Leftrightarrow\dfrac{2x+100}{95}+\dfrac{2x+100}{94}+\dfrac{2x+100}{93}=\dfrac{2x+100}{7}+\dfrac{2x+100}{6}+\dfrac{2x+100}{5}\)
\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}\right)=\left(2x+100\right)\left(\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{5}\right)\)
\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}\right)-\left(2x+100\right)\left(\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{5}\right)=0\)
\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\right)=0\)
mà \(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\ne0\)
nên 2x+100=0
\(\Leftrightarrow2x=-100\)
hay x=-50
Vậy: S={-50}
