Tìm min hoặc max : ` \frac{2(x^2+x+1)}{x^2+1}`

Tìm min hoặc max :
` \frac{2(x^2+x+1)}{x^2+1}`

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  1. Đáp án:

    $\min\left[\dfrac{2(x^2+ x+1)}{x^2+ 1}\right] = 1 \Leftrightarrow x = -1$

    $\max\left[\dfrac{2(x^2+ x+1)}{x^2+ 1}\right] = 3 \Leftrightarrow x = 1$

    Giải thích các bước giải:

    $\begin{array}{l}Đặt \,\,P =\dfrac{2(x^2+ x+1)}{x^2+ 1}\\ +) \quad P – 1 = \dfrac{2(x^2 + x + 1)}{x^2 + 1} – 1\\ \to P – 1 = \dfrac{2(x^2 + x +1) -(x^2 + 1)}{x^2 + 1}\\ \to P – 1 = \dfrac{x^2 + 2x + 1}{x^2 + 1}\\ \to P – 1 = \dfrac{(x+1)^2}{x^2+1} \geq 0\\ \to P – 1 \geq 0\\ \to P \geq 1\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x + 1 =0\Leftrightarrow x = -1\\ \to \min P = 1 \Leftrightarrow x = -1\\ +) \quad P -3 =\dfrac{2(x^2 + x + 1)}{x^2 + 1} – 3\\ \to P – 3= \dfrac{2(x^2 + x +1) -3(x^2 + 1)}{x^2 + 1}\\ \to P -3 = \dfrac{-x^2 + 2x – 1}{x^2 + 1}\\ \to P – 3 = -\dfrac{(x-1)^2}{x^2 +1} \leq 0\\ \to P – 3 \leq 0\\ \to P \leq 3\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x – 1 =0 \Leftrightarrow x = 1\\ \to \max P = 3 \Leftrightarrow x = 1 \end{array}$

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  2. $GTNN$

    `A=[2(x^2+x+1)]/(x^2+1)`

    `A=[(x^2+1)+(x^2+2x+1)]/(x^2+1)`

    `A=1+(x+1)^2/(x^2+1)≥1`

    Dấu `=` xảy ra `⇔(x+1)^2/(x^2+1)=0⇔x=-1`

    Vậy $Min_A=1⇔x=-1$

    $GTLN$

    `A=[2(x^2+x+1)]/(x^2+1)`

    `A=[3(x^2+1)-(x^2-2x+1)]/(x^2+1)`

    `A=3-(x-1)^2/(x^2+1)≤3`

    Dấu `=` xảy ra `⇔(x-1)^2/(x^2+1)=0⇔x=1`

    Vậy $Max_A=3⇔x=1$

     

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