Câu hỏi của La. Lousia

Question

$\left\{{}\begin{matrix}5x^2y-4xy^2+3y^3-2\left(x+y\right)=0\xy\left(x^2+y^2\right)+2=\left(x+y\right)^2\end{matrix}\right.$

Nguyễn Việt Lâm · Accepted Answer

$xy\left(x^2+y^2\right)+2=\left(x+y\right)^2$

$\Leftrightarrow xy\left[\left(x+y\right)^2-2xy\right]+2-\left(x+y\right)^2=0$

$\Leftrightarrow\left(x+y\right)^2\left(xy-1\right)-2\left[\left(xy\right)^2-1\right]=0$

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

$\Leftrightarrow\left(xy-1\right)\left[\left(x+y\right)^2-2xy-2\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}xy=1\x^2+y^2=2\end{matrix}\right.$

- Với $xy=1$

$xy\left(5x-4y\right)+3y^3-2x-2y=0$

$\Leftrightarrow3y^3+3x-6y=0$

$\Leftrightarrow3y^3+\frac{3}{y}-6y=0$

$\Leftrightarrow3y^4-6y^2+3=0\Leftrightarrow3\left(y^2-1\right)^2=0\Leftrightarrow...$

- Với $x^2+y^2=2$

$\Rightarrow2x^2y-4xy^2+3y\left(x^2+y^2\right)-2x-2y=0$

$\Leftrightarrow2x^2y-4xy^2-2x+4y=0$

$\Leftrightarrow2x\left(xy-1\right)-4y\left(xy-1\right)=0$

$\Leftrightarrow2\left(x-2y\right)\left(xy-1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=2y\xy=1\end{matrix}\right.$ $\Leftrightarrow...$Câu trả lời: $xy\left(x^2+y^2\right)+2=\left(x+y\right)^2$