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已知函数\(f\left ( {x} \right )\)满足\(f\left ( {x-\frac {1} {x}} \right )={x}^{2}+\frac {1} {{x}^{2}}\left ( {x≠0} \right )\),则\(f\left ( {x} \right )\)的解析式为( )
A.\(f\left ( {x} \right )=x+\frac {1} {x}\)
B.\(f\left ( {x} \right )={x}^{2}+2\)
C.\(f\left ( {x} \right )={x}^{2}\)
D.\(f\left ( {x} \right )=\left ( {x-\frac {1} {x}} \right )^{2}\)
参考答案:B
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