Question

giải hệ pt

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

Answer 1

- Với \(xy=0\) không phải nghiệm

- Với \(xy\ne0\) hệ tương đương

\(\left\{{}\begin{matrix}3x-2=\dfrac{1}{y^3}\\x^3+2=\dfrac{3}{y}\end{matrix}\right.\)

Đặt \(\dfrac{1}{y}=z\Rightarrow\left\{{}\begin{matrix}3x-2=z^3\\x^3+2=3z\end{matrix}\right.\)

\(\Rightarrow x^3+3x=z^3+3z\)

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

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

\(\Leftrightarrow x=z\)

Thế vào \(x^3+2=3z\Rightarrow x^3+2=3x\)

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

\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-2\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)