Question

Answer 1

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}x\left(x+y\right)+x=4\\\left(x+y\right)^2=4+\dfrac{5}{x^2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+1=\dfrac{4}{x}\\\left(x+y\right)^2=4+\dfrac{5}{x^2}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\\dfrac{1}{x}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+1=4v\\u^2=4+5v^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u=4v-1\\u^2=4+5v^2\end{matrix}\right.\)

\(\Rightarrow\left(4v-1\right)^2=4+5v^2\)

\(\Leftrightarrow11v^2-8v-3=0\Rightarrow\left\{{}\begin{matrix}v=1\Rightarrow u=3\\v=-\dfrac{3}{11}\Rightarrow u=-\dfrac{23}{11}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\dfrac{1}{x}=1\\x+y=3\end{matrix}\right.\\\left\{{}\begin{matrix}\dfrac{1}{x}=-\dfrac{3}{11}\\x+y=-\dfrac{23}{11}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x;y\)