hàm số

K

kakashi_hatake

$A(x_1, \ y_1), \ B(x_2, \ y_2)$

Giả sử $x_1>x_2$

Có $AB^2=(x_2-x_1)^2+(\dfrac{2}{x_2-1}-\dfrac{2}{x_1-1})^2=(x_1-x_2)^2.(1+\dfrac{4}{(x_1-1)^2(x_1-1)^2}) \\ (x_1-1)^2(x_1-1)^2 \le (\dfrac{x_1-1+1-x_2}{2})^4 \\ \rightarrow AB^2 \ge (x_1-x_2)^2+\dfrac{64}{(x_1-x_2)^2} \ge 16$

Dấu đẳng thức xảy ra khi $\begin{cases} x_1+x_2=2 \\ x_1-x_2=2\sqrt{2} \end{cases}$
 
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