## cho pt x^2-(m-1)x-m^2+m-2=0. Tìm m để pt có 2 nghiệm x1,x2sao cho x1^2-x2^2=3

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cho pt x^2-(m-1)x-m^2+m-2=0. Tìm m để pt có 2 nghiệm x1,x2sao cho x1^2-x2^2=3

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1 năm 2021-07-20T13:52:25+00:00 1 Answers 6 views 0

1. Đáp án: $m = 1,79$
$\begin{array}{l} {x^2} – \left( {m – 1} \right).x – {m^2} + m – 2 = 0\\ \Delta = {\left( {m – 1} \right)^2} – 4\left( { – {m^2} + m – 2} \right)\\ = {m^2} – 2m + 1 + 4{m^2} – 4m + 8\\ = 5{m^2} – 6m + 9 > 0 \end{array}$
$\begin{array}{l} Theo\,Viet:\left\{ \begin{array}{l} {x_1} + {x_2} = m – 1\\ {x_1}{x_2} = – {m^2} + m – 2 \end{array} \right.\\ x_1^2 – x_2^2 = 3\\ \Leftrightarrow \left( {{x_1} + {x_2}} \right)\left( {{x_1} – {x_2}} \right) = 3\\ \Leftrightarrow \left( {m – 1} \right).\sqrt {{{\left( {{x_1} – {x_2}} \right)}^2}} = 3\\ \Leftrightarrow \left( {m – 1} \right).\sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} – 4{x_1}{x_2}} = 3\\ \Leftrightarrow \left( {m – 1} \right).\sqrt {{{\left( {m – 1} \right)}^2} – 4\left( { – {m^2} + m – 2} \right)} = 3\\ \Leftrightarrow \left( {m – 1} \right).\sqrt {5{m^2} – 6m + 9} = 3\left( {dk:m > 1} \right)\\ \Leftrightarrow {\left( {m – 1} \right)^2}.\left( {5{m^2} – 6m + 9} \right) = 9\\ \Leftrightarrow \left( {{m^2} – 2m + 1} \right)\left( {5{m^2} – 6m + 9} \right) – 9 = 0\\ \Leftrightarrow 5{m^4} – 6{m^3} + 9{m^2} – 10{m^3} + 12{m^2} – 18m + 5{m^2} – 6m = 0\\ \Leftrightarrow 5{m^4} – 16{m^3} + 26{m^2} – 24m = 0\\ \Leftrightarrow m\left( {5{m^3} – 16{m^2} + 26m – 24} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} m = 0\left( {ktm} \right)\\ m = 1,79\left( {\left( {tm} \right)} \right) \end{array} \right.\\ Vậy\,m = 1,79 \end{array}$