Question

[Thái Viết Nam]

Giải hệ PT:

\(\left\{{}\begin{matrix}x^2-xy+y^2=3\left(x-y\right)\\x^2+xy+y^2=7\left(x-y\right)^3\end{matrix}\right.\)

Answer 1

[Nguyễn Việt Lâm]

\(\left\{{}\begin{matrix}\left(x-y\right)^2+xy=3\left(x-y\right)\\\left(x-y\right)^2+3xy=7\left(x-y\right)^3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3\left(x-y\right)^2+3xy=9\left(x-y\right)\\\left(x-y\right)^2+3xy=7\left(x-y\right)^3\end{matrix}\right.\)

\(\Rightarrow7\left(x-y\right)^3-9\left(x-y\right)=-2\left(x-y\right)^2\)

\(\Leftrightarrow7\left(x-y\right)^3+2\left(x-y\right)^2-9\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(7\left(x-y\right)^2+2\left(x-y\right)-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\x-y=1\\x-y=\dfrac{-9}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x\\y=x-1\\y=x+\dfrac{9}{7}\end{matrix}\right.\)

TH1: \(y=x\) thay vaò pt đầu:

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

TH2: \(y=x-1\) thay vào pt đầu:

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

\(\Rightarrow\left[{}\begin{matrix}x=2\Rightarrow y=1\\x=-1\Rightarrow y=-2\end{matrix}\right.\)

TH3: \(y=x+\dfrac{9}{7}\):

\(x^2-x\left(x+\dfrac{9}{7}\right)+\left(x+\dfrac{9}{7}\right)^2=\dfrac{-27}{7}\Leftrightarrow x^2+\dfrac{9}{7}x+\dfrac{270}{49}=0\) (vô nghiệm)

Vậy hệ đã cho có 3 cặp nghiệm:

\(\left(x;y\right)=\left(0;0\right);\left(2;1\right);\left(-1;-2\right)\)