C,D thuộc (P) nên \(C\left(x_1;x_1^2-4x_1-5\right);D\left(x_2;x_2^2-4x_2-5\right)\)
ABCD là hình bình hành
=>\(\overrightarrow{AB}=\overrightarrow{DC}\)
\(\overrightarrow{AB}=\left(3;-4\right);\overrightarrow{DC}=\left(x_1-x_2;x_1^2-4x_1-5-x_2^2+4x_2+5\right)\)
=>\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1^2-x_2^2-4x_1+4x_2=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_1-x_2=3\\\left(x_1-x_2\right)\left(x_1+x_2-4\right)=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2-4=-\dfrac{4}{x_1-x_2}=-\dfrac{4}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2=-\dfrac{4}{3}+4=\dfrac{12}{3}-\dfrac{4}{3}=\dfrac{8}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x_1=3+\dfrac{8}{3}=\dfrac{17}{3}\\x_1+x_2=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{17}{6}\\x_2=\dfrac{8}{3}-\dfrac{17}{6}=-\dfrac{1}{6}\end{matrix}\right.\)
Khi x=17/6 thì \(y=x^2-4x-5=\left(\dfrac{17}{6}\right)^2-4\cdot\dfrac{17}{6}-5=-\dfrac{299}{36}\)
Khi x=-1/6 thì \(y=\left(-\dfrac{1}{6}\right)^2-4\cdot\dfrac{-1}{6}-5=\dfrac{1}{36}+\dfrac{2}{3}-5=-\dfrac{155}{36}\)
Vậy: \(C\left(\dfrac{17}{6};-\dfrac{299}{36}\right);D\left(-\dfrac{1}{6};-\dfrac{155}{36}\right)\)
