http://diendantoanhoc.net/topic/151610-leftbeginmatrix-x33xy2-49-x2-8xyy28y-17x2-endmatrixright/

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nhân pt (2) vs 3 sau đó cộng pt (1) vs (2) ta đc

\(\left\{{}\begin{matrix}x^3+3xy^2=-46\\x^3+3xy^2+3x^2-24xy+3y^2=24y-51x-46\end{matrix}\right.\)

bây h ta chú ý tới pt dưới

\(x^3+3xy^2+3x^2-24xy+3y^2-24y+51x+46=0\)

\(\left(x+1\right)\left(x^2+2x+3y^2-24y+49\right)=0\)

\(\left(x+1\right)\left[\left(x+1\right)^2+3\left(y-4\right)^2\right]=0\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\x^3+3xy^2=-49\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\)

vậy hệ có 2 nghiệm

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+3xy^2=-49\\3x^2-24xy+3y^2=24y-51x\end{matrix}\right.\)

Cộng vế:

\(x^3+3x^2+3y^2\left(x+1\right)-24y\left(x+1\right)+51x+49=0\)

\(\Leftrightarrow\left(x+1\right)^3+3y^2\left(x+1\right)-24y\left(x+1\right)+48\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^3+3\left(x+1\right)\left(y-4\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2+3\left(y-2\right)^2\right]=0\)

Chỉ cần áp dụng cái giá trị tuyệt đối là rara

KHÓ QUÁ

x²-3xy+x=2y-2y²

<=>x²-3xy+2y²=2y-x

<=>(x-2y)(x-y)=2y-x

<=>(x-2y)(x-y+1)=0

đến đây thay vào pt 2 là ra