Question

Giải hệ pt: \(\left\{{}\begin{matrix}x^3\left(21y-20\right)=1\\x\left(y^3+20\right)=21\end{matrix}\right.\)

Answer 1

Nhận thấy \(x=0\) không phải nghiệm, hệ tương đương:

\(\left\{{}\begin{matrix}21y-20=\dfrac{1}{x^3}\\y^3+20=\dfrac{21}{x}\end{matrix}\right.\)

Cộng vế với vế:

\(y^3+21y=\dfrac{1}{x^3}+\dfrac{21}{x}\)

\(\Leftrightarrow y^3-\dfrac{1}{x^3}+21\left(y-\dfrac{1}{x}\right)=0\)

\(\Leftrightarrow\left(y-\dfrac{1}{x}\right)\left(y^2+\dfrac{y}{x}+\dfrac{1}{x^2}\right)+21\left(y-\dfrac{1}{x}\right)=0\)

\(\Leftrightarrow\left(y-\dfrac{1}{x}\right)\left(y^2+\dfrac{y}{x}+\dfrac{1}{x^2}+21\right)=0\)

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

\(\Leftrightarrow...\)