$\frac{1}{(x^2+2x+2)^2}$+ $\frac{1}{(x^2+2x+3)^2}$= $\frac{5}{4}$

Question

$\frac{1}{(x^2+2x+2)^2}$+ $\frac{1}{(x^2+2x+3)^2}$= $\frac{5}{4}$

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Reagan 17 phút 2021-10-24T17:57:38+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-24T17:58:57+00:00

    Đáp án: $S=\{-1\}$

     

    Giải thích các bước giải:

    $ĐKXĐ:x∈R$ 

    Ta có: 

    $(x^2+2x+2)^2=[(x+1)^2+1]^2≥1^2=1$

    `⇒\frac{1}{(x^2+2x+2)^2}≤\frac{1}{1}=1`

    $(x^2+2x+3)^2=[(x+1)^2+2]^2≥2^2=4$

    `⇒\frac{1}{(x^2+2x+3)^2}≤\frac{1}{4}`

    `⇒\frac{1}{(x^2+2x+2)^2}+\frac{1}{(x^2+2x+3)^2}≤1+\frac{1}{4}=\frac{5}{4}`

    Dấu bằng xảy ra

    $⇔(x+1)^2=0⇔x+1=0⇔x=-1$ (thỏa mãn)

    0
    2021-10-24T17:59:09+00:00

    Đáp án:(Chúc bạn học tốt!)

     

     

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