Đặt \(x^2+y^2=a;xy=b\)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}a^2-b^2=481\\a+b=37\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=481\\13\left(a+b\right)=481\end{matrix}\right.\)
Lấy pt trên trừ pt dưới suy ra \(\left(a+b\right)\left(a-b-13\right)=0\)
\(\Leftrightarrow37\left(a-b-13\right)=0\Rightarrow a=b+13\)
\(\Leftrightarrow x^2+y^2=xy+13\Leftrightarrow x^2-xy+y^2=13\)
\(\Rightarrow\frac{37}{13}\left(x^2-xy+y^2\right)=37\)
Từ đây kết hợp với pt thứ (2) của hệ ban đầu ta có: \(\frac{37}{13}\left(x^2-xy+y^2\right)=x^2+xy+y^2\)
\(\Leftrightarrow-\frac{2}{13}\left(3x-4y\right)\left(4x-3y\right)=0\)
Chị làm nốt thử ạ, em ko chắc đâu:v
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2\right)^2-x^2y^2=481\\\left(x+y\right)^2-xy=37\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=481\\\left(x+y\right)^2-xy=37\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a^2-2b\right)^2-b^2=481\\a^2-b=37\Rightarrow b=a^2-37\end{matrix}\right.\)
\(\Rightarrow\left(a^2-2\left(a^2-37\right)\right)^2-\left(a^2-37\right)^2=481\)
\(\Leftrightarrow\left(-a^2+74\right)^2-\left(a^2-37\right)^2-481=0\)
\(74a^2=3626\Rightarrow a^2=49\Rightarrow\left[{}\begin{matrix}a=7\Rightarrow b=12\\a=-7\Rightarrow b=12\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=7\\xy=12\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(4;3\right);\left(3;4\right)\)
\(\left\{{}\begin{matrix}x+y=-7\\xy=12\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-4;-3\right);\left(-3;-4\right)\)
