Question

giải hệ pt :

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

Answer 1

\(x-\dfrac{1}{x}=y-\dfrac{1}{y}\Leftrightarrow x-y+\dfrac{1}{y}-\dfrac{1}{x}=0\)

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

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

Thế vào pt dưới:

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x^2+x-1\right)=0\\x^4+x+2=0\Leftrightarrow\left(x^2-\dfrac{1}{2}\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}=0\left(vn\right)\end{matrix}\right.\)