Cho \[f\left( x \right) = \frac{1}{{{x^2} - 4x + 5}} - \frac{{{x^2}}}{4} + x\] Gọi \[M = \mathop {Max}\limits_{x \in \left[ {0;3} \right]} f(x);\;m = \mathop {Min}\limits_{x \in \left[ {0;3} \right]} f\left( x \right)\] Khi đó M−m bằng:

Ta có :

\[\begin{array}{*{20}{l}}{f\left( x \right) = \frac{1}{{{x^2} - 4x + 5}} - \frac{{{x^2}}}{4} + x}\\{f\left( x \right) = \frac{1}{{{x^2} - 4x + 5}} - \frac{{{x^2} - 4x}}{4}}\end{array}\]

Đặt\[t = {x^2} - 4x + 5\] với \[x \in \left[ {0;3} \right]\] ta có\[t' = 2x - 4 = 0 \Leftrightarrow x = 2 \in \left[ {0;3} \right]\]

Ta có \[t\left( 0 \right) = 5;\,\,t\left( 2 \right) = 1,\,\,t\left( 3 \right) = 2\]

 ⇒ Với\[x \in \left[ {0;3} \right]\]  thì\[t \in \left[ {1;5} \right]\]  khi đó hàm số trở thành\[f\left( t \right) = \frac{1}{t} - \frac{4}\] với\[t \in \left[ {1;5} \right]\]

Ta có\[f'\left( t \right) = - \frac{1}{{{t^2}}} - \frac{1}{4} < 0\,\,\forall t \in \left[ {1;5} \right]\]

⇒ Hàm số\[y = f\left( t \right)\] nghịch biến trên\[\left[ {1;5} \right]\]

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\mathop {max}\limits_{[0;3]} f(x) = \mathop {max}\limits_{[1;5]} f(t) = f(1) = 2 = M}\\{\mathop {min}\limits_{[0;3]} f(x) = \mathop {min}\limits_{[1;5]} f(t) = f(5) = \frac{1}{5} = m}\end{array}} \right.\)

Vậy \[M - m = 2 - \frac{1}{5} = \frac{9}{5}\]

Đáp án cần chọn là: D