Question

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

Answer 1

Với x=0 ko là nghiệm

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

\(\Rightarrow y^3+3y=\left(\frac{1}{x}\right)^3+3.\left(\frac{1}{x}\right)\)

\(\Rightarrow y^3-\left(\frac{1}{x}\right)^3+3\left(y-\frac{1}{x}\right)=0\)

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

\(\Leftrightarrow y=\frac{1}{x}\)

Thay vào pt đầu: \(x^3\left(2+\frac{3}{x}\right)=1\Leftrightarrow2x^3+3x^2-1=0\)

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