Đáp án:
$B.4$
Giải thích các bước giải:
Gọi $A \left( x_1 ; \dfrac{x_1}{x_1-2} \right), B \left( x_2 ; \dfrac{x_2}{x_2-2} \right)$
$\vec{AB} \left( x_2-x_1; \dfrac{x_2}{x_2-2} - \dfrac{x_1}{x_1-2} \right)$
$\to AB^2 = ( x_2-x_1)^2 + \left( \dfrac{x_2-x_1}{(x_2-2)(2-x_1)} \right)^2$
Áp dụng Cauchy
$\dfrac{(x_2-x_1)^2}{4} \ge (x_2-2)(2-x_1)$
$\to ( x_2-x_1)^2 + \left( \dfrac{x_2-x_1}{(x_2-2)(2-x_1)}\right)^2 \ge \sqrt{( x_2-x_1)^2. ( \dfrac{x_2-x_1}{(x_2-2)(2-x_1)}) ^2} \ge 2\sqrt{(x_1-x_2)^2.\dfrac{64}{(x_1-x_2)^2}} = 16$
Vậy $\text{min} = 4.$
