Đáp án:
$\min P = -\dfrac54 \Leftrightarrow x = -\dfrac52$
Giải thích các bước giải:
$\left(x \ne \dfrac{-5\pm \sqrt5}{2}\right)$
$\quad P = \dfrac{(x+1)(x+2)(x+3)(x+4) +1}{x^2 + 5x + 5}$
$\to P = \dfrac{[(x+1)(x+4)][(x+2)(x+3)] +1}{x^2 + 5x + 5}$
$\to P =\dfrac{(x^2 + 5x + 4)(x^2 + 5x + 6) +1}{x^2 + 5x + 5}$
$\to P =\dfrac{(x^2 + 5x + 5 -1)(x^2 + 5x + 5+1) +1}{x^2 + 5x + 5}$
$\to P = \dfrac{(x^2 + 5x+ 5)^2 - 1 + 1}{x^2 + 5x +5}$
$\to P = x^2 + 5x + 5$
$\to P = x^2 + 2\cdot\dfrac52x + \dfrac{25}{4} -\dfrac54$
$\to P = \left(x +\dfrac52\right)^2 -\dfrac54$
$\to P \geq -\dfrac54$
Dấu $=$ xảy ra $\Leftrightarrow x = -\dfrac52$
Vậy $\min P = -\dfrac54 \Leftrightarrow x = -\dfrac52$
