Help meeeeeeeeeeeeeeeeeeee

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Help meeeeeeeeeeeeeeeeeeee
help-meeeeeeeeeeeeeeeeeeee

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Trung Dũng 5 years 2021-05-18T01:19:44+00:00 1 Answers 17 views 0

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  1. yenoanh
    0
    2021-05-18T01:20:54+00:00
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    $y=\dfrac{1}{3}x^3+(m-1)x^2+(m^2-3m-4)x+2m-3\\ y’=x^2+2(m-1)x+m^2-3m-4\\ \Delta=(m-1)^2-m^2+3m+4\\ =m+5$

    Phương trình có hai nghiệm phân biệt $\Rightarrow m>-5$

    $Vi-et:x_1+x_2=-2(m-1)\\ x_1x_2=m^2-3m-4\\ M=x_1x_2-2(x_1+x_2)+10\\ =m^2-3m-4+4(m-1)+10\\ =m^2+m+2\\ =\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall \, m$

    Dấu “=” xảy ra $\Leftrightarrow m=\dfrac{1}{2}(TM)$

     

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