Cho\,\left\{ \begin{array}{l} a,b,c,d \in R\\ {a^2} + {b^2} + 4a - 2b + 4 = 0\left( 1 \right)\\ {c^2} + {d^2} - 6c - 2d + 6 = 0\left( 2 \right) \end{array} \right..\,\,Min,\,Max\,P = b + 3c + d - 2a - ac - bd = ?.
Giải:
\begin{array}{l} Do\,\left\{ \begin{array}{l} \left( 1 \right) \Leftrightarrow {\left( {a + 2} \right)^2} + {\left( {b - 1} \right)^2} = 1 \Rightarrow Coi\,M\left( {a;b} \right) \in \left( {{C_1}} \right):{\left( {x + 2} \right)^2} + {\left( {y - 1} \right)^2} = 1\\ \left( 2 \right) \Leftrightarrow {\left( {c - 3} \right)^2} + {\left( {d - 1} \right)^2} = 4 \Rightarrow Coi\,N\left( {c;d} \right) \in \left( {{C_2}} \right):{\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} = 4 \end{array} \right.\\ Do\,\left\{ \begin{array}{l} \left( 1 \right) \Leftrightarrow 2\left( {b - 2a} \right) = {a^2} + {b^2} + 4 \Leftrightarrow b - 2a = \dfrac{{{a^2} + {b^2} + 4}}{2}\\ \left( 2 \right) \Leftrightarrow 2\left( {3c + d} \right) = {c^2} + {d^2} + 6 \Leftrightarrow 3c + d = \dfrac{{{c^2} + {d^2} + 6}}{2} \end{array} \right.\\ \Rightarrow b + 3c + d - 2a = \dfrac{{{a^2} + {b^2} + {c^2} + {d^2}}}{2} + 5\\ \Rightarrow P = \dfrac{{{a^2} + {b^2} + {c^2} + {d^2}}}{2} - ac - bd + 5 = \dfrac{{{a^2} + {b^2} + {c^2} + {d^2} - 2ac - 2db}}{2} + 5\\ = \dfrac{{{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}}}{2} + 5 = \dfrac{{M{N^2}}}{2} + 5\\ \Rightarrow \left\{ \begin{array}{l} Min\,MN = {M_1}{N_1} = 2\left( {{M_1}( - 1;1)\& {N_1}(1;1)} \right) \Rightarrow Min\,P = \dfrac{4}{2} + 5 = 7 \Leftrightarrow a = - 1;b = c = d = 1\\ Max\,MN = {M_2}{N_2} = 8\left( {{M_2}( - 3;1)\& {N_2}(5;1)} \right) \Rightarrow Max\,P = \dfrac{{{8^2}}}{2} + 5 = 37 \Leftrightarrow a = - 3;b = 1;c = 5;d = 1\, \end{array} \right. \end{array}
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