Đáp án:
\[x = y = 1\]
Giải thích các bước giải:
ĐKXĐ: \(x \ne 0;\,\,y \ne 0\)
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + y + \dfrac{1}{x} + \dfrac{1}{y} = 4\\
{x^2} + {y^2} + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x + \dfrac{1}{x}} \right) + \left( {y + \dfrac{1}{y}} \right) = 4\\
\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) + \left( {{y^2} + \dfrac{1}{{{y^2}}}} \right) = 4
\end{array} \right.
\end{array}\)
Đặt \(\left\{ \begin{array}{l}
x + \dfrac{1}{x} = a\\
y + \dfrac{1}{y} = b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x^2} + 2.x.\dfrac{1}{x} + \dfrac{1}{{{x^2}}} = {a^2}\\
{y^2} + 2y.\dfrac{1}{y} + \dfrac{1}{{{y^2}}} = {b^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} + \dfrac{1}{{{x^2}}} = {a^2} - 2\\
{y^2} + \dfrac{1}{{{y^2}}} = {b^2} - 2
\end{array} \right.\)
Khi đó, hệ phương trình trên trở thành:
\(\begin{array}{l}
\left\{ \begin{array}{l}
a + b = 4\\
\left( {{a^2} - 2} \right) + \left( {{b^2} - 2} \right) = 4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = 4 - a\\
{a^2} + {b^2} = 8
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 4 - a\\
{a^2} + {\left( {4 - a} \right)^2} = 8
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = 4 - a\\
2{a^2} - 8a + 16 = 8
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 4 - a\\
{a^2} - 4a + 4 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = 4 - a\\
{\left( {a - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow a = b = 2
\end{array}\)
Do đó,
\(\left\{ \begin{array}{l}
x + \dfrac{1}{x} = 2\\
y + \dfrac{1}{y} = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{{x^2} + 1}}{x} = 2\\
\dfrac{{{y^2} + 1}}{y} = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 2x + 1 = 0\\
{y^2} - 2y + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right.x = y = 1\)
Vậy \(x = y = 1\)
