Đáp án:
b) \(m = - \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
C2:a)Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\\
{x_1}{x_2} = - 4
\end{array} \right.\\
B = \dfrac{{{x_1}^2 + {x_1}}}{{{x_2}}} + \dfrac{{{x_2}^2 + {x_2}}}{{{x_1}}}\\
= \dfrac{{{x_1}^3 + {x_1}^2 + {x_2}^3 + {x_2}^2}}{{{x_1}{x_2}}}\\
= \dfrac{{\left( {{x_1} + {x_2}} \right)\left( {{x_1}^2 - {x_1}{x_2} + {x_2}^2} \right) + \left( {{x_1}^2 + {x_2}^2} \right)}}{{{x_1}{x_2}}}\\
= \dfrac{{\left( {{x_1} + {x_2}} \right)\left( {{x_1}^2 - {x_1}{x_2} + {x_2}^2} \right) + \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right)}}{{{x_1}{x_2}}}\\
= \dfrac{{\left( {{x_1} + {x_2}} \right)\left( {{x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 3{x_1}{x_2}} \right) + \left( {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right)}}{{{x_1}{x_2}}}\\
= \dfrac{{\left( {{x_1} + {x_2}} \right)\left( {{{\left( {{x_1} + {x_2}} \right)}^2} - 3{x_1}{x_2}} \right) + \left( {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right)}}{{{x_1}{x_2}}}\\
= \dfrac{{ - 2\left( {4 - 3.\left( { - 4} \right)} \right) + \left( {4 - 2.\left( { - 4} \right)} \right)}}{{ - 4}}\\
= 5
\end{array}\)
b) Để phương trình có nghiệm
\(\begin{array}{l}
\to \Delta \ge 0\\
\to 4{m^2} - 4m + 1 - 4\left( { - 2m - 1} \right) \ge 0\\
\to 4{m^2} - 4m + 1 + 8m + 4 \ge 0\\
\to 4{m^2} + 4m + 5 \ge 0\left( {ld} \right)\forall m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 1\\
{x_1}{x_2} = - 2m - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_1} + {x_2} + {x_1}{x_2} = - 2\\
{x_1} + {x_2} = 2m - 1
\end{array} \right.\\
{x_1}{x_2} + {x_2}^2 + {x_1}{x_2}^2 - 2{x_1} = 4\\
\to {x_1}{x_2} + {x_2}^2 + {x_1}{x_2}^2 + \left( {{x_1} + {x_2} + {x_1}{x_2}} \right){x_1} = 4\\
\to {x_1}{x_2} + {x_2}^2 + {x_1}{x_2}^2 + {x_1}^2 + {x_1}{x_2} + {x_1}^2{x_2} = 4\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 + {x_1}{x_2}\left( {{x_1} + {x_2}} \right) = 4\\
\to {\left( {{x_1} + {x_2}} \right)^2} + {x_1}{x_2}\left( {{x_1} + {x_2}} \right) = 4\\
\to 4{m^2} - 4m + 1 + \left( { - 2m - 1} \right)\left( {2m - 1} \right) = 4\\
\to 4{m^2} - 4m + 1 - 4{m^2} + 1 = 4\\
\to - 4m = 2\\
\to m = - \dfrac{1}{2}
\end{array}\)
