\(20.x^2+px+1=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-p\\ab=1\end{matrix}\right.\)
\(x^2+qx+2=0\Leftrightarrow\left\{{}\begin{matrix}b+c=-q\\bc=2\end{matrix}\right.\)
\(\Rightarrow pq=\left(-p\right)\left(-q\right)=\left(a+b\right)\left(b+c\right)\)
\(\Rightarrow A=\left(a+b\right)\left(b+c\right)-\left(b-a\right)\left(b-c\right)=ab+ac+b^2+bc-b^2+bc+ab-ac=2ab+2bc=2+2.2=6\)
\(c19:\Rightarrow ADHE\) \(là\) \(hình\) \(chữ\) \(nhật\)(mình hướng dẫn thôi bn tự làm)
\(\Rightarrow\left\{{}\begin{matrix}AB^2+AC^2=100\\\dfrac{1}{4^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\end{matrix}\right.\)\(giải\) \(hệ\Rightarrow AB,AC\)
\(dùng:AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=......\)
\(\Rightarrow CH=BC-BH=.....\)
\(\dfrac{1}{DH^2}=\dfrac{1}{BH^2}+\dfrac{1}{AH^2}\Rightarrow DH=....\)
\(\dfrac{1}{HE^2}=\dfrac{1}{AH^2}+\dfrac{1}{HC^2}\Rightarrow HE=,,,,,\Rightarrow S\left(ADHE\right)=DH.HE=....\)
