Tìm $min_{P}=\dfrac{(x+1)(x+2)(x+3)(x+4)+1}{x^{2}+5x+5}$

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1. 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 $\min P = -\dfrac54 \Leftrightarrow x = -\dfrac52$

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2. `P=((x+1)(x+2)(x+3)(x+4)+1)/(x^2+5x+5)`

   `⇒P=([(x+1)(x+4)][(x+2)(x+3)+1])/(x^2+5x+5)`

   `⇒P=((x^2+5x+4)(x^2+5x+6)+1)/(x^2+5x+5)` $(*) $

   Đặt $x^2+5x+5=a$ thay vào $(*)$ ta có:

   `⇒P=((a-1)(a+1)+1)/a`

   `⇒P=(a^2-1+1)/a`

   `⇒P=a^2/a`

   `⇒P=a`

   `⇒P=x^2+5x+5`

   `⇒P=x^2+2.5/2x+25/4+5-25/4`

   `⇒P=(x+5/2)^2-5/4≥-5/4`

   Dấu “=” xảy ra khi `x+5/2=0`

                             `⇒x=-5/2`

   Vậy `Pmin=-5/4` đạt tại `x=-5/2`

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