Cho phương trình như sau \(a{x^2} + bx + c = 0\;\;\left( {a \ne 0} \right)\) có hai nghiệm \({x_1},\;{x_2}\) thỏa mãn \(0 \le {x_1} \le {x_2} \le 2.\) Tìm giá trị nhỏ nhất của biểu thức \(L = \dfrac{{3{a^2} - ab + ac}}{{5{a^2} - 3ab + {b^2}}}.\)

Phương trình có hai nghiệm \({x_1},\;{x_2}\) thỏa mãn \(0 \le {x_1} \le {x_2} \le 2\)

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

Theo hệ thức Vi-ét ta có: \(\left\{ \begin{array}{l}{x_1} + {x_2} =  - \dfrac{b}{a}\\{x_1}{x_2} = \dfrac{c}{a}\end{array} \right..\)

Theo đề bài ta có:

\(\begin{array}{l}L = \dfrac{{3{a^2} - ab + ac}}{{5{a^2} - 3ab + {b^2}}} \\\;\;= \dfrac{{3 - \dfrac{b}{a} + \dfrac{c}{a}}}{{5 - 3.\dfrac{b}{a} + {{\left( {\dfrac{b}{a}} \right)}^2}}}\;\;\left( {do\;\;a \ne 0} \right)\\\;\; = \dfrac{{3 + \left( {{x_1} + {x_2}} \right) + {x_1}{x_2}}}{{5 + 3\left( {{x_1} + {x_2}} \right) + {{\left( {{x_1} + {x_2}} \right)}^2}}}\;\;\;\left( {L > 0\;\;\forall \;0 \le {x_1} \le {x_2} \le 2} \right)\\\;\; = \dfrac{{3 + {x_1} + {x_2} + {x_1}{x_2}}}{{5 + 3{x_1} + 3{x_2} + x_1^2 + x_2^2 + 2{x_1}{x_2}}}.\\ \Rightarrow \dfrac{1}{L} = \dfrac{{5 + 3{x_1} + 3{x_2} + x_1^2 + x_2^2 + 2{x_1}{x_2}}}{{3 + {x_1} + {x_2} + {x_1}{x_2}}}.\end{array}\)

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

\(\begin{array}{l} \Rightarrow \dfrac{1}{L} \le \dfrac{{5 + 3{x_1} + 3{x_2} + 2{x_1} + 2{x_2} + 2{x_1}{x_2}}}{{3 + {x_1} + {x_2} + {x_1}{x_2}}}\\\;\;\;\;\;\;\; = \dfrac{{5 + 5{x_1} + 5{x_2} + 2{x_1}{x_2}}}{{3 + {x_1} + {x_2} + {x_1}{x_2}}}\\\;\;\;\;\;\;\; = \dfrac{{3{x_1}{x_2} + 3{x_1} + 3{x_2} + 9 - {x_1}{x_2} + 2{x_1} + 2{x_2} - 4}}{{3 + {x_1} + {x_2} + {x_1}{x_2}}}\\\;\;\;\;\;\;\; = \dfrac{{3\left( {3 + {x_1} + {x_2} + {x_1}{x_2}} \right) - \left( {{x_2} - 2} \right){x_1} + 2\left( {{x_2} - 2} \right)}}{{3 + {x_1} + {x_2} + {x_1}{x_2}}}\\\;\;\;\;\;\;\; = 3 - \dfrac{{\left( {{x_2} - 2} \right)\left( {{x_1} - 2} \right)}}{{3 + {x_1} + {x_2} + {x_1}{x_2}}} \le 3\;\;\;\left( {do\;\;\left( {{x_2} - 2} \right)\left( {{x_1} - 2} \right) \ge 0} \right)\\ \Rightarrow 0 \le \dfrac{1}{L} \le 3 \Leftrightarrow 3L \ge 1 \Leftrightarrow L \ge \dfrac{1}{3}\\ \Rightarrow Min\;L = \dfrac{1}{3}.\end{array}\)

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 \(Min\;L = \dfrac{1}{3}\) khi \(\left( {{x_1};\;{x_2}} \right) = \left\{ {\left( {0;\;2} \right),\;\left( {2;\;0} \right),\;\left( {2;\;2} \right)} \right\}.\)