giải hệ phương trình $\left \{ {{x^{2}+x+ \frac{1}{y} (1+\frac{1}{y} )=4} \atop { x^{3}+\frac{x}{y^{2}}+\frac{x^{2}}{y} +\frac{1}{y^{3}} =4}} \rig

Đáp án:

\[x = y = 1\]

Giải thích các bước giải:

 ĐKXĐ: \(y \ne 0\)

Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + x + \frac{1}{y}\left( {1 + \frac{1}{y}} \right) = 4\\
{x^3} + \frac{x}{{{y^2}}} + \frac{{{x^2}}}{y} + \frac{1}{{{y^2}}} = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + x + \frac{1}{y} + \frac{1}{{{y^2}}} = 4\\
x\left( {{x^2} + \frac{1}{{{y^2}}}} \right) + \frac{1}{y}\left( {{x^2} + \frac{1}{{{y^2}}}} \right) = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {{x^2} + \frac{1}{{{y^2}}}} \right) + \left( {x + \frac{1}{y}} \right) = 4\\
\left( {{x^2} + \frac{1}{{{y^2}}}} \right)\left( {x + \frac{1}{y}} \right) = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + \frac{1}{{{y^2}}} = 4 – \left( {x + \frac{1}{y}} \right)\\
\left( {4 – \left( {x + \frac{1}{y}} \right)} \right)\left( {x + \frac{1}{y}} \right) = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + \frac{1}{{{y^2}}} = 4 – \left( {x + \frac{1}{y}} \right)\\
{\left( {x + \frac{1}{y}} \right)^2} – 4\left( {x + \frac{1}{y}} \right) + 4 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + \frac{1}{y} = 2\\
{x^2} + \frac{1}{{{y^2}}} = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{y} = 2 – x\\
{x^2} + {\left( {2 – x} \right)^2} = 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{y} = 2 – x\\
{x^2} – 2x + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
\frac{1}{y} = 1
\end{array} \right. \Rightarrow x = y = 1
\end{array}\)