$\frac{x^{2}-10x-29 }{1971}$+$\frac{x^{2}-10x-27}{1973}$=$\frac{x^{2}-10x-1971}{29}$ +$\frac{x^{2}-10x-1973}{24}$
$\frac{x^{2}-10x-29 }{1971}$+$\frac{x^{2}-10x-27}{1973}$=$\frac{x^{2}-10x-1971}{29}$ +$\frac{x^{2}-10x-1973}{24}$
Bạn tham khảo câu trả lời của mình:
⇔$(\frac{x^2-10x-29}{1971}-1)+$ $(\frac{x^2-10x-27}{1973}-1)-$ $(\frac{x^2-10x-1971}{29}-1)-$ $(\frac{x^2-10x-1973}{27}-1)=0$ <=>$\frac{x^2-10-2000}{1971}+$ $\frac{x^2-10x-2000}{1973}-$ $\frac{x^2-10x-2000}{29}-$ $\frac{x^2-10x-200}{27}=0$
<=>$(x^2-10x-2000)($$\frac{1}{1971}+$ $\frac{1}{1973}-$ $\frac{1}{29}-$ $\frac{1}{27})=0$
⇔$x^2-20x-2000=0$ (Vì: $\frac{1}{1971}+$ $\frac{1}{1973}-$ $\frac{1}{29}-$ $\frac{1}{27})$ ∦0
⇔(x+40)(x-50)=0
⇔\(\left[ \begin{array}{l}x=-40\\x=50\end{array} \right.\)
Vậy S={-40;50}