\(\left\{{}\begin{matrix}x^2+y^2-x-y=102\\xy+x+y=69\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy-\left(x+y\right)=102\\xy+\left(x+y\right)=69\end{matrix}\right.\)
Đặt \(S=x+y\)
\(P=xy\)
\(\Rightarrow\left\{{}\begin{matrix}S^2-2P-S=102\\P+S=69\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S^2-2.\left(69-S\right)-S=102\\P=69-S\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S^2-138+2S-S=102\\P=69-S\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S^2+S-240\\P=69-S\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}S=15\\S=-16\end{matrix}\right.\\P=69-S\end{matrix}\right.\)
+) Với \(S=15;P=54\) có :
\(\left\{{}\begin{matrix}S=x+y\\P=xy\end{matrix}\right.\Rightarrow x,y\) là nghiệm của pt : \(x^2-15x+54=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\Rightarrow y=6\\x=6\Rightarrow y=9\end{matrix}\right.\)
+) Với \(S=-16;P=85\) có :
\(\left\{{}\begin{matrix}S=x+y\\P=xy\end{matrix}\right.\Rightarrow x,y\) là nghiệm của pt : \(x^2+16x+85=0\)
\(\Leftrightarrow\left(x+8\right)^2+21=0\) (vô lí)
\(\Rightarrow\) pt vô nghiệm
Vậy nghiệm của hệ pt đã cho là \(\left(x;y\right):\left(6;9\right),\left(9;6\right)\)
\(2\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=1\\\left(x+y\right)\left(x^2-xy+y^2\right)-xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=1\\\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-xy=0\end{matrix}\right.\)
Đặt \(S=x+y;P=xy\)
\(\Rightarrow\left\{{}\begin{matrix}S^2-P=1\\S\left(S^2-3P\right)-P=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S^2-P=1\\S^3-3PS-P=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}P=S^2-1\\S^3-3S\left(S^2-1\right)+1-S^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}P=S^2-1\\S^3-3S^3+3S-S^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2S^3-S^2+3S+1=0\\P=S^2-1\end{matrix}\right.\)
Còn lại thì bấm máy tính !!!
