1/ Cho PT : $x^{2}$ – (3m+1)x +2 $m^{2}$ +m =0 tìm m để ║x1 – x2 ║=1 2/ Cho PT $x^{2}$ – (m+2)x +m -1 =0 Đặt A = $x1^{2}$ + $x2^{2}$ – 4x1x2. T

0 bình luận về “1/ Cho PT : $x^{2}$ – (3m+1)x +2 $m^{2}$ +m =0 tìm m để ║x1 – x2 ║=1 2/ Cho PT $x^{2}$ – (m+2)x +m -1 =0 Đặt A = $x1^{2}$ + $x2^{2}$ – 4x1x2. T”

1. Đáp án:

   $\begin{array}{l}
   1){x^2} – \left( {3m + 1} \right)x + 2{m^2} + m = 0\\
    \Rightarrow \Delta  > 0\\
    \Rightarrow {\left( {3m + 1} \right)^2} – 4\left( {2{m^2} + m} \right) > 0\\
    \Rightarrow 9{m^2} + 6m + 1 – 8{m^2} – 4m > 0\\
    \Rightarrow {m^2} + 2m + 1 > 0\\
    \Rightarrow m \ne  – 1\\
   Theo\,Viet:\left\{ \begin{array}{l}
   {x_1} + {x_2} = 3m + 1\\
   {x_1}{x_2} = 2{m^2} + m
   \end{array} \right.\\
   \left| {{x_1} – {x_2}} \right| = 1\\
    \Rightarrow {\left( {{x_1} – {x_2}} \right)^2} = 1\\
    \Rightarrow {\left( {{x_1} + {x_2}} \right)^2} – 4{x_1}{x_2} = 1\\
    \Rightarrow {\left( {3m + 1} \right)^2} – 4\left( {2{m^2} + m} \right) = 1\\
    \Rightarrow {m^2} + 2m + 1 = 1\\
    \Rightarrow m\left( {m + 2} \right) = 0\\
    \Rightarrow \left[ \begin{array}{l}
   m = 0\left( {tm} \right)\\
   m =  – 2\left( {tm} \right)
   \end{array} \right.\\
   2){x^2} – \left( {m + 2} \right).x + m – 1 = 0\\
    \Rightarrow \Delta  > 0\\
    \Rightarrow {\left( {m + 2} \right)^2} – 4m + 4 > 0\\
    \Rightarrow {m^2} + 4m + 4 – 4m + 4 > 0\\
    \Rightarrow {m^2} + 8 > 0\left( {tm} \right)\\
   Theo\,Viet:\left\{ \begin{array}{l}
   {x_1} + {x_2} = m + 2\\
   {x_1}{x_2} = m – 1
   \end{array} \right.\\
   A = x_1^2 + x_2^2 – 4{x_1}{x_2}\\
    = \left( {{x_1} + {x_2}} \right) – 6{x_1}{x_2}\\
    = {\left( {m + 2} \right)^2} – 6.\left( {m – 1} \right)\\
    = {m^2} + 4m + 4 – 6m + 6\\
    = {m^2} – 2m + 1 + 9\\
    = {\left( {m – 1} \right)^2} + 9 \ge 9\\
    \Rightarrow GTNN:A = 9\\
   Khi:m = 1
   \end{array}$

   Vậy m=1

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