Cho phương trình \(a{{x}^{2}}+bx+c=0\ \ \left( a\n...

D

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Phương trình có hai nghiệm \({{x}_{1}},\ {{x}_{2}}\) thỏa mãn \(0\le {{x}_{1}}\le {{x}_{2}}\le 2\)

\( \Leftrightarrow \left\{ \begin{array}{l}\Delta  \ge 0\\af\left( 0 \right) \ge 0\\af\left( 2 \right) \ge 0\\\frac{S}{2} > 0\\\frac{S}{2} < 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{b^2} - 4ac > 0\\ac \ge 0\\a\left( {4a + 2b + c} \right) \ge 0\\ - \frac{b}{{2a}} > 0\\ - \frac{b}{{2a}} < 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{b^2} \ge 4ac\\ac \ge 0\\a\left( {4a + 2b + c} \right) \ge 0\\\frac{b}{{2a}} < 0\\\frac{{4a + b}}{{2a}} > 0\end{array} \right..\)

Theo hệ thức Vi-ét ta có: \(\left\{ \begin{align}  {{x}_{1}}+{{x}_{2}}=-\frac{b}{a} \\  {{x}_{1}}{{x}_{2}}=\frac{c}{a} \\ \end{align} \right..\)

Theo đề bài ta có:

\(\begin{align}  L=\frac{3{{a}^{2}}-ab+ac}{5{{a}^{2}}-3ab+{{b}^{2}}}=\frac{3-\frac{b}{a}+\frac{c}{a}}{5-3.\frac{b}{a}+{{\left( \frac{b}{a} \right)}^{2}}}\ \ \left( do\ \ a\ne 0 \right) \\  \ \ =\frac{3+\left( {{x}_{1}}+{{x}_{2}} \right)+{{x}_{1}}{{x}_{2}}}{5+3\left( {{x}_{1}}+{{x}_{2}} \right)+{{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}}\ \ \ \left( L>0\ \ \forall \ 0\le {{x}_{1}}\le {{x}_{2}}\le 2 \right) \\  \ \ =\frac{3+{{x}_{1}}+{{x}_{2}}+{{x}_{1}}{{x}_{2}}}{5+3{{x}_{1}}+3{{x}_{2}}+x_{1}^{2}+x_{2}^{2}+2{{x}_{1}}{{x}_{2}}}. \\  \Rightarrow \frac{1}{L}=\frac{5+3{{x}_{1}}+3{{x}_{2}}+x_{1}^{2}+x_{2}^{2}+2{{x}_{1}}{{x}_{2}}}{3+{{x}_{1}}+{{x}_{2}}+{{x}_{1}}{{x}_{2}}}. \\ \end{align}\)

Vì 

\(0 \le {x_1} \le {x_2} \le 2 \Rightarrow \left\{ \begin{array}{l}x_1^2 \le 2{x_1}\\x_2^2 \le 2{x_2}\\{x_1} - 2 \le 0\\{x_2} - 2 \le 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x_1^2 + x_2^2 \le 2{x_1} + 2{x_2}\\\left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) \ge 0\end{array} \right..\)

\(\begin{align}  \Rightarrow \frac{1}{L}\le \frac{5+3{{x}_{1}}+3{{x}_{2}}+2{{x}_{1}}+2{{x}_{2}}+2{{x}_{1}}{{x}_{2}}}{3+{{x}_{1}}+{{x}_{2}}+{{x}_{1}}{{x}_{2}}}=\frac{5+5{{x}_{1}}+5{{x}_{2}}+2{{x}_{1}}{{x}_{2}}}{3+{{x}_{1}}+{{x}_{2}}+{{x}_{1}}{{x}_{2}}} \\  \ \ \ \ \ \ \ =\frac{3{{x}_{1}}{{x}_{2}}+3{{x}_{1}}+3{{x}_{2}}+9-{{x}_{1}}{{x}_{2}}+2{{x}_{1}}+2{{x}_{2}}-4}{3+{{x}_{1}}+{{x}_{2}}+{{x}_{1}}{{x}_{2}}} \\  \ \ \ \ \ \ \ =\frac{3\left( 3+{{x}_{1}}+{{x}_{2}}+{{x}_{1}}{{x}_{2}} \right)-\left( {{x}_{2}}-2 \right){{x}_{1}}+2\left( {{x}_{2}}-2 \right)}{3+{{x}_{1}}+{{x}_{2}}+{{x}_{1}}{{x}_{2}}} \\  \ \ \ \ \ \ \ =3-\frac{\left( {{x}_{2}}-2 \right)\left( {{x}_{1}}-2 \right)}{3+{{x}_{1}}+{{x}_{2}}+{{x}_{1}}{{x}_{2}}}\le 3\ \ \ \left( do\ \ \left( {{x}_{2}}-2 \right)\left( {{x}_{1}}-2 \right)\ge 0 \right) \\  \Rightarrow 0\le \frac{1}{L}\le 3\Leftrightarrow 3L\ge 1\Leftrightarrow L\ge \frac{1}{3} \\  \Rightarrow Min\ L=\frac{1}{3}. \\ \end{align}\)

Dấu “=” xảy ra

\( \Leftrightarrow \left\{ \begin{array}{l}x_1^2 = 2{x_1}\\x_2^2 = 2{x_2}\\\left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_1}\left( {{x_1} - 2} \right) = 0\\{x_2}\left( {{x_2} - 2} \right) = 0\\\left[ \begin{array}{l}{x_1} - 2 = 0\\{x_2} - 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x_1} = 2\\{x_2} = 2\end{array} \right.\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}{x_1} = 0\\{x_1} = 2\end{array} \right.\\\left[ \begin{array}{l}{x_2} = 0\\{x_2} = 2\end{array} \right.\\\left[ \begin{array}{l}{x_1} = 2\\{x_2} = 2\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{x_1} = 0\\{x_2} = 2\end{array} \right.\\\left\{ \begin{array}{l}{x_1} = 2\\{x_2} = 0\end{array} \right.\\\left\{ \begin{array}{l}{x_1} = 2\\{x_2} = 2\end{array} \right.\end{array} \right..\)

Vậy \(Min\ L=\frac{1}{3}\ \ khi\ \ \left( {{x}_{1}};\ {{x}_{2}} \right)=\left\{ \left( 0;\ 2 \right),\ \left( 2;\ 0 \right),\ \left( 2;\ 2 \right) \right\}.\)