## 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

Question

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. Tìm m sao cho A đạt giá trị nhỏ nhất

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1 năm 2021-10-09T03:05:11+00:00 1 Answers 5 views 0

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