Cho pt x^2-2m+1x+m^2+2=0. tìm m để pt có 2 nghiệm x1,x2 tm hệ thức 3x1x2-5×1+x2+7=0 27/11/2021 Bởi Maria Cho pt x^2-2m+1x+m^2+2=0. tìm m để pt có 2 nghiệm x1,x2 tm hệ thức 3x1x2-5×1+x2+7=0
Đáp án: $\begin{array}{l}{x^2} – \left( {2m + 1} \right).x + {m^2} + 2 = 0\\DK:\Delta \ge 0\\ \Rightarrow {\left( {2m + 1} \right)^2} – 4\left( {{m^2} + 2} \right) \ge 0\\ \Rightarrow 4{m^2} + 4m + 1 – 4{m^2} – 8 \ge 0\\ \Rightarrow 4m \ge 7\\ \Rightarrow m \ge \dfrac{7}{4}\\Theo\,Viet:\left\{ \begin{array}{l}{x_1} + {x_2} = 2m + 1\\{x_1}{x_2} = {m^2} + 2\end{array} \right.\\3{x_1}{x_2} – 5\left( {{x_1} + {x_2}} \right) + 7 = 0\\ \Rightarrow 3.\left( {{m^2} + 2} \right) – 5\left( {2m + 1} \right) + 7 = 0\\ \Rightarrow 3{m^2} + 6 – 10m – 5 + 7 = 0\\ \Rightarrow 3{m^2} – 10m + 8 = 0\\ \Rightarrow 3{m^2} – 6m – 4m + 8 = 0\\ \Rightarrow \left( {m – 2} \right)\left( {3m – 4} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}m = 2\left( {tm} \right)\\m = \dfrac{4}{3}\left( {ktm} \right)\end{array} \right.\\Vậy\,m = 2\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
{x^2} – \left( {2m + 1} \right).x + {m^2} + 2 = 0\\
DK:\Delta \ge 0\\
\Rightarrow {\left( {2m + 1} \right)^2} – 4\left( {{m^2} + 2} \right) \ge 0\\
\Rightarrow 4{m^2} + 4m + 1 – 4{m^2} – 8 \ge 0\\
\Rightarrow 4m \ge 7\\
\Rightarrow m \ge \dfrac{7}{4}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 1\\
{x_1}{x_2} = {m^2} + 2
\end{array} \right.\\
3{x_1}{x_2} – 5\left( {{x_1} + {x_2}} \right) + 7 = 0\\
\Rightarrow 3.\left( {{m^2} + 2} \right) – 5\left( {2m + 1} \right) + 7 = 0\\
\Rightarrow 3{m^2} + 6 – 10m – 5 + 7 = 0\\
\Rightarrow 3{m^2} – 10m + 8 = 0\\
\Rightarrow 3{m^2} – 6m – 4m + 8 = 0\\
\Rightarrow \left( {m – 2} \right)\left( {3m – 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 2\left( {tm} \right)\\
m = \dfrac{4}{3}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,m = 2
\end{array}$