\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
\(ĐKXĐ:\) \(x\ne2009\) \(;\) \(x\ne2010\)
Đặt \(a=x-2010\) (với \(a\ne0\) ), ta được:
\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\) \(\Leftrightarrow\) \(\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\) \(\Leftrightarrow\) \(49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow\) \(8a^2+8a-30=0\) \(\Leftrightarrow\) \(4a^2+4a-15=0\) \(\Leftrightarrow\) \(\left(2a+1\right)^2-16=0\)
\(\Leftrightarrow\) \(\left(2a-3\right)\left(2a+5\right)=0\) \(\Leftrightarrow\) \(^{a=\frac{3}{2}}_{a=-\frac{5}{2}}\) ( thỏa mãn điều kiện )
Do đó, \(x=\frac{4023}{2}\) hoặc \(x=\frac{4015}{2}\) (thỏa mãn \(ĐKXĐ\) )
Vậy, \(S=\left\{\frac{4023}{2};\frac{4015}{2}\right\}\)
