Toán Lớp 9: Cho pt: x² – (m + 3)x + m + 2 = 0
Định m để pt có 2 nghiệm phân biệt $x_1$,$x_2$ thỏa: $x_1^{4}$ + $x_2^{4}$ = 626
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Toán Lớp 9: Cho pt: x² – (m + 3)x + m + 2 = 0
Định m để pt có 2 nghiệm phân biệt $x_1$,$x_2$ thỏa: $x_1^{4}$ + $x_2^{4}$ = 626
Comments ( 2 )
m = 3\\
m = – 7
\end{array} \right.\)
\to {m^2} + 6m + 9 – 4\left( {m + 2} \right) > 0\\
\to {m^2} + 6m + 9 – 4m – 8 > 0\\
\to {m^2} + 2m + 1 > 0\\
\to {\left( {m + 1} \right)^2} > 0\\
\to m \ne – 1\\
Vi – et:\left\{ \begin{array}{l}
{x_1} + {x_2} = m + 3\\
{x_1}{x_2} = m + 2
\end{array} \right.\\
{x_1}^4 + {x_2}^4 = 626\\
\to {x_1}^4 + {x_2}^4 + 2{x_1}^2{x_2}^2 – 2{x_1}^2{x_2}^2 = 626\\
\to {\left( {{x_1}^2 + {x_2}^2} \right)^2} – 2{\left( {{x_1}{x_2}} \right)^2} = 626\\
\to {\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} – 2{x_1}{x_2}} \right)^2} – 2{\left( {{x_1}{x_2}} \right)^2} = 626\\
\to {\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} – 2{x_1}{x_2}} \right]^2} – 2{\left( {{x_1}{x_2}} \right)^2} = 626\\
\to {\left( {{m^2} + 6m + 9 – 2\left( {m + 2} \right)} \right)^2} – 2{\left( {m + 2} \right)^2} = 626\\
\to {\left( {{m^2} + 4m + 5} \right)^2} – 2{m^2} – 8m – 8 = 626\\
\to {m^4} + 16{m^2} + 25 + 8{m^3} + 10{m^2} + 40m – 2{m^2} – 8m – 8 = 626\\
\to {m^4} + 8{m^3} + 24{m^2} + 32m – 609 = 0\\
\to {m^4} – 3{m^3} + 11{m^3} – 33{m^2} + 57{m^2} – 171m + 203m – 609 = 0\\
\to {m^3}\left( {m – 3} \right) + 11{m^2}\left( {m – 3} \right) + 57m\left( {m – 3} \right) + 203\left( {m – 3} \right) = 0\\
\to \left( {m – 3} \right)\left( {{m^3} + 11{m^2} + 57m + 203} \right) = 0\\
\to \left[ \begin{array}{l}
m = 3\\
m = – 7
\end{array} \right.
\end{array}\)