Tìm x , y biết rằng x ^2 + y^2 + 1/ x^2 + 1 / y^2 = 4
Ta có \({x^2} + {y^2} + \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} = 4\)
\({x^2} + \frac{1}{{{x^2}}} + {y^2} + \frac{1}{{{y^2}}} - 4 = 0\)
\(\left( {{x^2} - 2 + \frac{1}{{{x^2}}}} \right) + \left( {{y^2} - 2 + \frac{1}{{{y^2}}}} \right) = 0\)
\({\left( {x - \frac{1}{x}} \right)^2} + {\left( {y - \frac{1}{y}} \right)^2} = 0\)
Ta thấy\({\left( {x - \frac{1}{x}} \right)^2} \ge 0\,;\,\,{\left( {y - \frac{1}{y}} \right)^2} \ge 0\).
Để \({\left( {x - \frac{1}{x}} \right)^2} + {\left( {y - \frac{1}{y}} \right)^2} = 0\) thì \(\left\{ \begin{array}{l}{\left( {x - \frac{1}{x}} \right)^2} = 0\\{\left( {y - \frac{1}{y}} \right)^2} = 0\end{array} \right.\) nên \(\left\{ \begin{array}{l}x - \frac{1}{x} = 0\\y - \frac{1}{y} = 0\end{array} \right.\) do đó \(\left\{ \begin{array}{l}{x^2} - 1 = 0\\{y^2} - 1 = 0\end{array} \right..\)
Ta có bảng sau:
\(x\) | 1 | 1 | \( - 1\) | \( - 1\) |
\(y\) | 1 | \( - 1\) | 1 | \( - 1\) |
Vậy các cặp \(\left( {x\,;\,\,y} \right)\) thỏa mãn biểu thức là \(\left( {1\,;\,\,1} \right)\,;\,\,\left( { - 1\,;\,\,1} \right)\,;\,\,\left( {1\,;\,\, - 1} \right)\,;\,\,\left( { - 1\,;\,\, - 1} \right)\,\).