Đặt \(S=x+y;P=xy\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}S\left(1-P\right)=5P\left(3\right)\\S^2\left(1+\dfrac{1}{P^2}\right)=9\left(4\right)\end{matrix}\right.\)
\(\left(3\right)\Rightarrow S=\dfrac{5P}{1-P}\)
\(\left(4\right)\Rightarrow\left(\dfrac{5P}{1-P}\right)^2.\left(1+\dfrac{1}{P^2}\right)=9\)
\(\Leftrightarrow\dfrac{25P^2}{\left(1-P\right)^2}.\dfrac{1+P^2}{P^2}=9\)
\(\Leftrightarrow\dfrac{25+25P^2}{\left(1-P\right)^2}=9\)
\(\Leftrightarrow25+25P^2=9-18P+9P^2\)
\(\Leftrightarrow16P^2+18P+16=0\)
\(\Leftrightarrow8P^2+9P+8=0\left(5\right)\)
\(\Delta=81-256=-175< 0\Rightarrow\left(5\right)VN\)
Vậy HPT cho vô nghiệm
