$x^{2}$ -mx-4=0 MinA = $\frac{2(x1 + x2 ) +7}{x1^{2}+x2^{2} }$

Question

$x^{2}$ -mx-4=0
MinA = $\frac{2(x1 + x2 ) +7}{x1^{2}+x2^{2} }$

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Harper 1 năm 2021-09-20T14:00:48+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-09-20T14:02:13+00:00

    Đáp án:

    Để pt có 2 nghiệm x1 và x2 thì:

    $\begin{array}{l}
    \Delta  \ge 0\\
     \Rightarrow {m^2} – 4.\left( { – 4} \right) \ge 0\\
     \Rightarrow {m^2} + 16 \ge 0\left( {luon\,dung} \right)\\
    Theo\,Viet:\left\{ \begin{array}{l}
    {x_1} + {x_2} = m\\
    {x_1}{x_2} =  – 4
    \end{array} \right.\\
    A = \dfrac{{2\left( {{x_1} + {x_2}} \right) + 7}}{{x_1^2 + x_2^2}}\\
     = \dfrac{{2.m + 7}}{{{{\left( {{x_1} + {x_2}} \right)}^2} – 2{x_1}{x_2}}}\\
     = \dfrac{{2m + 7}}{{{m^2} – 2.\left( { – 4} \right)}}\\
     = \dfrac{{2m + 7}}{{{m^2} + 8}}\\
     \Rightarrow A.\left( {{m^2} + 8} \right) = 2m + 7\\
     \Rightarrow A.{m^2} – 2m + 8A – 7 = 0\\
     \Rightarrow \left\{ \begin{array}{l}
    a = A\\
    b =  – 2\\
    c = 8A – 7
    \end{array} \right.\\
     \Rightarrow \Delta ‘ \ge 0\\
     \Rightarrow {\left( { – 1} \right)^2} – A.\left( {8A – 7} \right) \ge 0\\
     \Rightarrow 1 – 8{A^2} + 7A \ge 0\\
     \Rightarrow 8{A^2} – 7A – 1 \le 0\\
     \Rightarrow \left( {8A + 1} \right)\left( {A – 1} \right) \le 0\\
     \Rightarrow  – \dfrac{1}{8} \le A \le 1\\
     \Rightarrow \min A =  – \dfrac{1}{8} \Leftrightarrow m =  – 8
    \end{array}$

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