Đáp án:
\(P=3.\)
Giải thích các bước giải:
\(\begin{array}{l} y = f\left( x \right) = {x^3} + 3x + 1\,\,\,\,\,\left( {{C_1}} \right)\\ y = g\left( x \right) = {x^3} - 3{x^2} + 6x - 1\,\,\,\left( {{C_2}} \right)\\ \text{Gọi }M\left( {x;\,\,y} \right) \in \left( {{C_1}} \right)\\ M'\left( {x';\,\,y'} \right)\text{ là ảnh của }M\text{ qua }{T_{\overrightarrow v }}\\ \Rightarrow \left\{ \begin{array}{l} x' = x + a\\ y' = y + b \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = x' - a\\ y = y' - b \end{array} \right.\\ \Rightarrow M\left( {x' - a;\,\,y' - b} \right) \in \left( {{C_1}} \right)\\ \Rightarrow y' - b = {\left( {x' - a} \right)^3} + 3\left( {x' - a} \right) + 1\\ \Leftrightarrow y' - b = x{'^3} - 3ax{'^2} + 3{a^2}x' - {a^3} + 3x' - 3a + 1\\ \Leftrightarrow y' = x{'^3} - 3ax{'^2} + \left( {3{a^2} + 3} \right)x' - {a^3} -3a+ b + 1\\ \Rightarrow \left\{ \begin{array}{l} - 3a = - 3\\ 3{a^2} + 3 = 6\\ - {a^3} -3a+ b + 1 = - 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 1\\ {a^2} = 1\\ b - {a^3}-3a= - 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 1\\ b =-2+a^2.a+3a=-2+1+3= 2 \end{array} \right.\\ \Rightarrow P = a + b = 1 +2 = 3. \end{array}\)
