a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
nên \(x^2-10x-2000=0\)
\(\Leftrightarrow x^2+40x-50x-2000=0\)
\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy: S={-40;50}
