Toán $x^{2}$ -mx-4=0 MinA = $\frac{2(x1 + x2 ) +7}{x1^{2}+x2^{2} }$ 20/09/2021 By Harper $x^{2}$ -mx-4=0 MinA = $\frac{2(x1 + x2 ) +7}{x1^{2}+x2^{2} }$
Đáp án: Để pt có 2 nghiệm x1 và x2 thì: $\begin{array}{l}\Delta \ge 0\\ \Rightarrow {m^2} – 4.\left( { – 4} \right) \ge 0\\ \Rightarrow {m^2} + 16 \ge 0\left( {luon\,dung} \right)\\Theo\,Viet:\left\{ \begin{array}{l}{x_1} + {x_2} = m\\{x_1}{x_2} = – 4\end{array} \right.\\A = \dfrac{{2\left( {{x_1} + {x_2}} \right) + 7}}{{x_1^2 + x_2^2}}\\ = \dfrac{{2.m + 7}}{{{{\left( {{x_1} + {x_2}} \right)}^2} – 2{x_1}{x_2}}}\\ = \dfrac{{2m + 7}}{{{m^2} – 2.\left( { – 4} \right)}}\\ = \dfrac{{2m + 7}}{{{m^2} + 8}}\\ \Rightarrow A.\left( {{m^2} + 8} \right) = 2m + 7\\ \Rightarrow A.{m^2} – 2m + 8A – 7 = 0\\ \Rightarrow \left\{ \begin{array}{l}a = A\\b = – 2\\c = 8A – 7\end{array} \right.\\ \Rightarrow \Delta ‘ \ge 0\\ \Rightarrow {\left( { – 1} \right)^2} – A.\left( {8A – 7} \right) \ge 0\\ \Rightarrow 1 – 8{A^2} + 7A \ge 0\\ \Rightarrow 8{A^2} – 7A – 1 \le 0\\ \Rightarrow \left( {8A + 1} \right)\left( {A – 1} \right) \le 0\\ \Rightarrow – \dfrac{1}{8} \le A \le 1\\ \Rightarrow \min A = – \dfrac{1}{8} \Leftrightarrow m = – 8\end{array}$ Trả lời
Đáp án:
Để pt có 2 nghiệm x1 và x2 thì:
$\begin{array}{l}
\Delta \ge 0\\
\Rightarrow {m^2} – 4.\left( { – 4} \right) \ge 0\\
\Rightarrow {m^2} + 16 \ge 0\left( {luon\,dung} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = – 4
\end{array} \right.\\
A = \dfrac{{2\left( {{x_1} + {x_2}} \right) + 7}}{{x_1^2 + x_2^2}}\\
= \dfrac{{2.m + 7}}{{{{\left( {{x_1} + {x_2}} \right)}^2} – 2{x_1}{x_2}}}\\
= \dfrac{{2m + 7}}{{{m^2} – 2.\left( { – 4} \right)}}\\
= \dfrac{{2m + 7}}{{{m^2} + 8}}\\
\Rightarrow A.\left( {{m^2} + 8} \right) = 2m + 7\\
\Rightarrow A.{m^2} – 2m + 8A – 7 = 0\\
\Rightarrow \left\{ \begin{array}{l}
a = A\\
b = – 2\\
c = 8A – 7
\end{array} \right.\\
\Rightarrow \Delta ‘ \ge 0\\
\Rightarrow {\left( { – 1} \right)^2} – A.\left( {8A – 7} \right) \ge 0\\
\Rightarrow 1 – 8{A^2} + 7A \ge 0\\
\Rightarrow 8{A^2} – 7A – 1 \le 0\\
\Rightarrow \left( {8A + 1} \right)\left( {A – 1} \right) \le 0\\
\Rightarrow – \dfrac{1}{8} \le A \le 1\\
\Rightarrow \min A = – \dfrac{1}{8} \Leftrightarrow m = – 8
\end{array}$