Question

[Hi Mn]

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

Answer 1

[Trần Tuấn Hoàng]

- Đk: \(xy\ne0\)

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

\(\left(1\right)\Rightarrow\left(x+\dfrac{1}{y}\right)^2+\left(x+\dfrac{1}{y}\right)-2.\dfrac{x}{y}=4\)

\(\left(2\right)\Rightarrow\left(x+\dfrac{1}{y}\right)\left(x^2-\dfrac{x}{y}+\dfrac{1}{y^2}\right)+\dfrac{x}{y}\left(x+\dfrac{1}{y}\right)=4\)

\(\Rightarrow\left(x+\dfrac{1}{y}\right)\left(x^2+\dfrac{1}{y^2}\right)=4\)

\(\Rightarrow\left(x+\dfrac{1}{y}\right)\left[\left(x+\dfrac{1}{y}\right)^2-2.\dfrac{x}{y}\right]=4\)

\(\Rightarrow\left(x+\dfrac{1}{y}\right)^3-2\left(x+\dfrac{1}{y}\right).\dfrac{x}{y}=4\)

Đặt \(m=x+\dfrac{1}{y};n=\dfrac{x}{y}\left(m,n\ne0\right)\). Khi đó ta có:

\(\left\{{}\begin{matrix}m^2+m-2n=4\left(3\right)\\m^3-2mn=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-2n=4-m\\m\left(m^2-2n\right)=4\end{matrix}\right.\)

\(\Rightarrow m\left(4-m\right)=4\)

\(\Leftrightarrow m^2-4m+4=0\)

\(\Leftrightarrow\left(m-2\right)^2=0\)

\(\Leftrightarrow m=2\). Thay vào (3) ta được:

\(2^2+2-2n=4\)

\(\Leftrightarrow n=1\)

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

\(\Rightarrow x,\dfrac{1}{y}\) là 2 nghiệm của phương trình \(X^2-2X+1\).

\(\Delta=\left(-2\right)^2-4.1.1=0\)

\(\Rightarrow\)Phương trình có nghiệm kép \(X_{1,2}=\dfrac{2}{2}=1\)

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

Vậy hệ đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(1;1\right)\)