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1. \(\begin{cases}x+y+xy\left(2x+y\right)=5xy\\x+y+xy\left(3x-y\right)=4xy\end{cases}\) \(\Leftrightarrow\begin{cases}2y-x=1\\x+y+xy\left(2x+y\right)=5xy\end{cases}\) (trừ 2 vế cho nhau)
\(\Leftrightarrow\begin{cases}x=2y-1\\\left(2y-1\right)+y+\left(2y-1\right)y\left(4y-2+y\right)=5\left(2y-1\right)y\end{cases}\) \(\Leftrightarrow\begin{cases}x=2y-1\\10y^3-19y^2+10y-1=0\end{cases}\) \(\Leftrightarrow\begin{cases}x=1\\y=1\end{cases}\)
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đặt \(\left\{{}\begin{matrix}S=X+Y\\P=X.Y\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}S+P=5\\S^2-P=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}P=5-S\\S^2+S-12=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}P=5-S\\\left[{}\begin{matrix}S=-4\\S=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}S=-4\\P=9\end{matrix}\right.\\\left\{{}\begin{matrix}S=3\\P=2\end{matrix}\right.\end{matrix}\right.\)
suy ra tìm đc x và y
b,c tương tự