\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-2xy=10\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\) ta được:
\(\left\{{}\begin{matrix}a+b=7\\a^2-2b=10\end{matrix}\right.\) \(\Rightarrow a^2-2\left(7-a\right)-10=0\)
\(\Rightarrow a^2+2a-24=0\Rightarrow\left[{}\begin{matrix}a=4\Rightarrow b=3\\a=-6\Rightarrow b=13\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) theo Viet đao, x và y là nghiệm:
\(t^2-4t+3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)
Đặt x+y=a và xy=b
=> hệ pt <=> \(\left\{{}\begin{matrix}a+b=7\\a^2-2b=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a+2b=14\\a^2-2b=10\end{matrix}\right.\)
\(\Leftrightarrow a^2+2a=24\)
\(\Leftrightarrow a^2-4a+6a-24=0\)
\(\Leftrightarrow a\left(a-4\right)+6\left(a-4\right)=0\)
\(\Leftrightarrow\left(a-4\right)\left(a+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)
TH1: a=4 => b= 3
Hay \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4-y\\\left(4-y\right)y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4-y\\4y-y^2-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4-y\\y^2-4y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4-y\\\left(y-1\right)\left(y-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4-y\\\left[{}\begin{matrix}y=1\\y=3\end{matrix}\right.\end{matrix}\right.\)
+) với y=1 => x= 3
+) Với y=3 => x=1
TH2: giải tương tự
