Question

giải hệ pt

\(\left\{{}\begin{matrix}2xy+y+2=-8x\\x^2y^2+xy+1=7x^2\end{matrix}\right.\)

Answer 1

Lời giải:

Ta có:

\(\left\{\begin{matrix} 2xy+y+2=-8x\\ x^2y^2+xy+1=7x^2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} 2(xy+1)=-(8x+y)\\ (xy+1)^2=7x^2+xy\end{matrix}\right.\)

\(\Rightarrow \left[\frac{-(8x+y)}{2}\right]^2=7x^2+xy\)

\(\Leftrightarrow 64x^2+y^2+16xy=28x^2+4xy\)

\(\Leftrightarrow 36x^2+y^2+12xy=0\)

\(\Leftrightarrow (6x+y)^2=0\Rightarrow y=-6x\)

Thay vào pt đầu tiên suy ra:

\(-6x^2+x+1=0\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\rightarrow y=-3\\ x=\frac{-1}{3}\Rightarrow y=2\end{matrix}\right.\)

Vậy...........