Question

giải hệ sau :

x^2 *(y+1)=6y-2

x^4*y^2 +2x^2*y^2 +y(x^2 +1)=12y^2 -1

 

Answer 1

\(\left\{{}\begin{matrix}x^2\left(y+1\right)=6y-2\\x^4y^2+2x^2y^2+y\left(x^2+1\right)=12y^2-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x^2\left(y+1\right)=6y-2\\\left(x^2y\right)^2+2x^2yy+yx^2=12y^2-y-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x^2=\dfrac{6y-2}{y+1}\\y^2=x^4\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{7\pm\sqrt{41}}{2}\\x=\sqrt{\dfrac{7\pm\sqrt{41}}{2}}\end{matrix}\right.\)