Question

[Lê Đình Quân]

\(\left\{{}\begin{matrix}x^2+x+\frac{1}{y}\left(1+\frac{1}{y}\right)=4\\x^3+\frac{x}{y^2}+\frac{x^2}{y}+\frac{1}{y^3}=4\end{matrix}\right.\)

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: ...

\(\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+x+\frac{1}{y}=4\\x^2\left(x+\frac{1}{y}\right)+\frac{1}{y^2}\left(x+\frac{1}{y}\right)=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+x+\frac{1}{y}=4\\\left(x^2+\frac{1}{y^2}\right)\left(x+\frac{1}{y}\right)=4\end{matrix}\right.\)

Đặt \(\left(x^2+\frac{1}{y^2};x+\frac{1}{y}\right)=\left(u;v\right)\Rightarrow\left\{{}\begin{matrix}u+v=4\\uv=4\end{matrix}\right.\)

Theo Viet đảo, u và v là nghiệm:

\(t^2-4t+4=0\Rightarrow t=2\Rightarrow\left\{{}\begin{matrix}x^2+\frac{1}{y^2}=2\\x+\frac{1}{y}=2\end{matrix}\right.\)

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