Question

giải hệ phương trình:

1, \(\left\{{}\begin{matrix}x^2\left(1+y^2\right)=2\\1+xy+x^2y^2=3x^2\end{matrix}\right.\)

2, \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(y+1\right)^2=27xy\\\left(x^2+1\right)\left(y^2+1\right)=10xy\end{matrix}\right.\)

Answer 1

1) \(\left\{{}\begin{matrix}x^2\left(1+y^2\right)=2\\1+xy+x^2y^2=3x^2\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2 +x^2y^2=2\\1+xy+x^2y^2-3x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-3x^2+xy+x^2y^2=-1\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-3\left(2-x^2y^2\right)+xy+x^2y^2=-1\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-6+3x^2y^2+xy+x^2y^2=-1\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-5+4x^2y^2+xy=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\-5+4x^2y^2+xy=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\\left(4xy+5\right)\left(xy-1\right)=0\end{matrix}\right.\)

Ta có 2 trường hợp:

TH1: \(\left\{{}\begin{matrix}x^2=2-x^2y^2\\4xy+5=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2=2-x^2y^2\\xy=\dfrac{-5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=2-\dfrac{25}{16}\\xy=\dfrac{-5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{7}}{4}\\y=\dfrac{-5}{\sqrt{7}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x^2=2-x^2y^2\\xy=1\end{matrix}\right.\)\(\left\{{}\begin{matrix}x^2=2-1\\xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\xy=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Answer 2

Câu 2 mình chưa nghĩ ra. Sorry bạn nheee