+ Đặt: $\left \{ {{t = x + \frac{1}{y}} \atop {v = \frac{1}{x}}} \right.$ ⇒$tu = (x + \frac{1}{x})(1 + \frac {1}{y}) = xy = \frac{1}{xy} + 2$

+ Ta có hệ phương trình: 

CHÚC EM HỌC TỐT. XIN HAY NHẤT.

Đáp án:

$-$

Giải thích các bước giải:

$\left[ \begin{array}{l}t+u=\frac{9}{2}1/4+3/2tu-2\end{array} \right.$

`=>` $\left[ \begin{array}{l}2t+2y=9\\-4tu+6t+9=0\end{array} \right. $

`=>` $\left[ \begin{array}{l}2u=9-2t\\-2t(9-2t)+6t+9=0\end{array} \right. $

`=>` $\left[ \begin{array}{l}2u=9-2t\\4t^2-126t+9=0\end{array} \right. $

`=>` $\left[ \begin{array}{l}2u=9-2t\\(2t-3)^2=0\end{array} \right. $

`=>` $\left[ \begin{array}{l}2u=9-2t\\2t=3\end{array} \right. $

`=>` $\left[ \begin{array}{l}u=3\\t=3/2\end{array} \right.$

`=>` $\left[ \begin{array}{l}x+1/y=3/2\\y+1/x=3\end{array} \right.$

`=>` $\left[ \begin{array}{l}xy-3/2y+1=0\\xy-3x+1=0\end{array} \right. $

`=>` $\left[ \begin{array}{l}-3/2y+3x=0\\xy-3x+1=0\end{array} \right. $

`=>` $\left[ \begin{array}{l}y=2x\\2x^2-3x+1=0\end{array} \right. $

`=>` $\left[ \begin{array}{l}y=2x\\(x-1)(2x-1)=0\end{array} \right. $

`=>` $\left[ \begin{array}{l}x=1\\y=2\end{array} \right. $

$=$ $\left[ \begin{array}{l}x=1/2\\y=1\end{array} \right. $

Vậy $x$ là $1$ , $y$ là$2$

       $x$ là `1/2` , $y$ là $1$