}$\text{Giải thích các bước giải:}$
$\text{Ta có :}$
$\dfrac{1909-x}{91} + \dfrac{1907-x}{93} + \dfrac{1905-x}{95} + \dfrac{1903-x}{97} = -4$
$⇒ (\dfrac{1909-x}{91} + \dfrac{1907-x}{93} + \dfrac{1905-x}{95} + \dfrac{1903-x}{97}) + 4 = 0$
$⇒ (\dfrac{1909-x}{91} + 1) + (\dfrac{1907-x}{93} + 1) + (\dfrac{1905-x}{95} + 1) + (\dfrac{1903-x}{97} + 1) = 0$
$⇒ \dfrac{2000-x}{91} + \dfrac{2000-x}{93} + \dfrac{2000-x}{95} + \dfrac{2000-x}{97} = 0$
$⇒ (2000 - x)(\dfrac{1}{91} + \dfrac{1}{93} + \dfrac{1}{95} + \dfrac{1}{97}) = 0$
$⇒ 2000 - x = 0 (\dfrac{1}{91} + \dfrac{1}{93} + \dfrac{1}{95} + \dfrac{1}{97} \neq 0)$
$⇒ x = 2000$
$\text{Vậy x = 2000}$
$\text{Chúc bạn học tốt !}$
