Câu hỏi của Nguyen

Question

Giải hpt:

$\left\{{}\begin{matrix}xy\left(4xy+y+4\right)=y^2\left(2y+5\right)-1\2xy\left(x-2y\right)+x-14y=0\end{matrix}\right.$

Help me Nguyễn Việt Lâm , Akai Haruma

Kiêm Hùng · Accepted Answer

$\left(1\right)2xy\left(x-2y\right)+x-14y=0$

$\Leftrightarrow2xy\left(x-2y\right)+\left(x-2y\right)-12y=0$

​$\Leftrightarrow\left(2xy+1\right)\left(x-2y\right)=12y$​

$\left(2\right)xy\left(4xy+y+4\right)=y^2\left(2y+5\right)-1$

$\Leftrightarrow4x^2y^2+x^2y+4xy=2y^3+5y^2-1$

$\Leftrightarrow4x^2y^2+x^2y+4xy-2y^3-5y^2+1=0$

$\Leftrightarrow4x^2y^2+8xy+1-4xy+x^2+4y^2+x^2y-x^2-2y^3+2y^2-11y^2=0$

$\Leftrightarrow\left(2xy+1\right)^2+\left(x-2y\right)^2+x^2\left(y-1\right)-2y^2\left(y-1\right)=11y^2$

$\Leftrightarrow\left(2xy+1\right)^2+\left(x-2y\right)^2+\left(x^2-2y^2\right)\left(y-1\right)=11y^2$

_ Phân tích được tới đây :)_Câu trả lời: $\left(1\right)2xy\left(x-2y\right)+x-14y=0$