\begin{array}{l} Cho\,\left\{ \begin{array}{l} a,b,c,d \in R\\ a + 2b = 9;\,c + 2d = 4 \end{array} \right..\,Min\,P = \sqrt {{a^2} + {b^2} - 12a - 8b + 52} + \sqrt {{a^2} + {b^2} + {c^2} + {d^2} - 2ac - 2bd} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \sqrt {{c^2} + {d^2} - 4c + 8d + 20} = ? \end{array}.
Giải:
Do\,P = \sqrt {{{\left( {a - 6} \right)}^2} + {{\left( {b - 4} \right)}^2}} + \sqrt {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}} + \sqrt {{{\left( {c - 2} \right)}^2} + {{\left( {d + 4} \right)}^2}} .
\begin{array}{*{20}{l}} {Coi{\mkern 1mu} \left\{ {\begin{array}{*{20}{l}} {M\left( {a;b} \right) \Rightarrow M \in {d_1}:{f_1}\left( {x,y} \right) = x + 2y - 9 = 0}\\ {N\left( {c;d} \right) \Rightarrow N \in {d_2}:{f_2}\left( {x,y} \right) = x + 2y = 4}\\ {A\left( {6;4} \right),{\mkern 1mu} D\left( {2; - 4} \right)} \end{array}} \right.}\\ { \Rightarrow \left\{ {\begin{array}{*{20}{l}} {P = AM + MN + ND}\\ {{d_1}//{d_2}} \end{array}} \right.}\\ {Do{\mkern 1mu} \left\{ {\begin{array}{*{20}{l}} {{f_1}\left( A \right) = 6 + 8 - 9 = 5 > 0}\\ {{f_2}\left( A \right) = 6 + 8 - 4 = 10 > 0} \end{array}} \right.} \end{array}.
Khi đó A nằm phía trên 2 đường thẳng {d_1}\& {d_2}.
Mặt khác, Do\,\left\{ \begin{array}{l} {f_1}\left( D \right) = 2 - 8 - 9 = - 15 < 0\\ {f_2}\left( D \right) = 2 - 8 - 4 = - 10 < 0 \end{array} \right.
Khi đó D nằm phía dưới 2 đường thẳng {d_1}\& {d_2}. \begin{array}{l} \Rightarrow P\,\min \Leftrightarrow \left\{ \begin{array}{l} AM\,\min \Leftrightarrow AM = AB = d\left( {A \to {d_1}} \right)\left( {AB \bot {d_1}} \right) \Leftrightarrow M \equiv B\\ MN\,\min \Leftrightarrow MN = d\left( {{d_1};{d_2}} \right) = BC\\ ND\,\min \Leftrightarrow ND = CD = d\left( {D \to {d_2}} \right)\left( {CD \bot {d_2}} \right) \Leftrightarrow N \equiv D \end{array} \right. \Rightarrow AB \equiv BC \equiv CD \equiv AD\\ \Rightarrow Min\,P = d\left( {A \to {d_1}} \right) + d\left( {{d_1};{d_2}} \right) + d\left( {D \to {d_2}} \right) = \frac{5}{{\sqrt 5 }} + \frac{5}{{\sqrt 5 }} + \frac{{10}}{{\sqrt 5 }} = 4\sqrt 5 \\ \Rightarrow AD:2x - y = 8 \Rightarrow \left\{ \begin{array}{l} B = AD \cap {d_1}:\left\{ \begin{array}{l} 2x - y = 8\\ x + 2y = 9 \end{array} \right. \Leftrightarrow B\left( {5;2} \right)\\ C = AD \cap {d_2}:\left\{ \begin{array}{l} 2x - y = 8\\ x + 2y = 4 \end{array} \right. \Leftrightarrow C\left( {4;0} \right) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = 5\\ b = 2\\ c = 4\\ d = 0 \end{array} \right. \end{array}
Một cách giải khác hay hơn, các bạn tham khảo nhé!
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