\(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(x+y\right)+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)
đặc (x + y) là a ; xy là b
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=7\\a^2-b=13\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=7-b\\\left(7-b\right)^2-b=13\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=7-b\\49-14b+b^2-b=13\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=7-b\\b^2-15b+36=0\end{matrix}\right.\) \(\left\{{}\begin{matrix}a=7-b\\\left\{{}\begin{matrix}b=12\\b=3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}b=12\\a=-5\end{matrix}\right.\\\left\{{}\begin{matrix}b=3\\a=4\end{matrix}\right.\end{matrix}\right.\)
vậy : (x + y) = -5 ; xy = 12 và (x + y) = 4 ; xy = 3
vậy x ; y lần lược là nghiệm của các phương trình sau
(1) : x2 + 5x + 12 = 0 (vô nghiệm)
(2) : x2 - 4x + 3 = 0 \(\Rightarrow\) \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\end{matrix}\right.\)
vậy \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)
Đặt x + y = a ; xy = b
=> (1) trở thành: \(\left\{{}\begin{matrix}a+b=7\left(2\right)\\a^2-b=13\left(3\right)\end{matrix}\right.\)
Cộng (2) với (3) ta được: a2 + a = 20
<=> a2 + a - 20 = 0
<=> (a - 4)(a + 5) = 0
<=> \(\left[{}\begin{matrix}a=4\\a=-5\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=-5\\b=12\end{matrix}\right.\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-5\\xy=12\end{matrix}\right.\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4-y\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5-y\\xy=12\end{matrix}\right.\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\left(4-y\right)y=3\\-\left(5+y\right)y=12\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}y^2-4y+3=0\\y^2+5y+12=0\end{matrix}\right.\)
<=> y2 - 4y + 3 = 0 ( vì y2 + 5y + 12 = 0 vô nghiệm)
<=> ( y - 1)(y - 3) = 0
<=> \(\left[{}\begin{matrix}y=1\\y=3\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\end{matrix}\right.\)
Vậy ( x;y) = ( 3;1), (1;3)
