Question

\(\left\{{}\begin{matrix}x^2+2y^2+3xy=12\\\left(x+y\right)\left(2+xy\right)=12\end{matrix}\right.\)

Answer 1

(1) \(\Leftrightarrow x^2+2y^2+2xy+xy=12\Leftrightarrow x\left(x+y\right)+2y\left(x+y\right)=12\Leftrightarrow\left(x+y\right)\left(2y+x\right)=12\)Lay (1) - (2) \(\Rightarrow\left(x+y\right)\left(x+2y-2-xy\right)=0\Leftrightarrow\left(x+y\right)\left(x-2\right)\left(1-y\right)=0\)

*) \(x=-y\)=> ko tm vi (2) khac 0

*) x=2 thay vao 1

\(4+2y^2+6y=12\Leftrightarrow2y^2+6y-8=0\Leftrightarrow y^2+3y-4=0\)

=> y1=1 ; y2=-4

*) y=1

thay vao 1

\(\Rightarrow x^2+2+3x=12\Leftrightarrow x^2+3x-10=0\)

=> x1=2 ; x2=-5