Question

\(\left\{{}\begin{matrix}x^2+y^2+2x+4y=8\\\left(x+2y+1\right)\left(9+3y^2+4xy\right)=64\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+4y^2+2x+4y+4xy+1=9+3y^2+4xy\\\left(x+2y+1\right)\left(9+3y^2+4xy\right)=64\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+2y+1\right)^2=9+3y^2+4xy\\\left(x+2y+1\right)\left(9+3y^2+4xy\right)=64\end{matrix}\right.\)

=>(x+2y+1)^3=64

=>x+2y+1=4

=>x=3-2y

x=3-2y vào x^2+y^2+2x+4y=8, ta được:

(3-2y)^2+y^2+2(3-2y)+4y=8

=>y=1 hoặc y=7/5

=>x=1 hoặc x=1/5