Đáp án: $ F\le -4$
Giải thích các bước giải:
Ta có:
$F=\dfrac{5-3x}{3-x}+\dfrac{2x-5}{x-2}$
$\to F=\dfrac{3(3-x)-4}{3-x}+\dfrac{2(x-2)-1}{x-2}$
$\to F=3-\dfrac{4}{3-x}+2-\dfrac{1}{x-2}$
$\to F=5-(\dfrac{2^2}{3-x}+\dfrac{1^2}{x-2})$
Mà $\dfrac{2^2}{3-x}+\dfrac{1^2}{x-2}\ge \dfrac{(2+1)^2}{3-x+x-2}$
$\to \dfrac{2^2}{3-x}+\dfrac{1^2}{x-2}\ge 9$
$\to 5-(\dfrac{2^2}{3-x}+\dfrac{1^2}{x-2})\le -4$
$\to F\le -4$
Dấu = xảy ra khi $\dfrac{2}{3-x}=\dfrac{1}{x-2}\to x=\dfrac73$
