已知0<x<1,化简√{[x-(1/x)]^2+4}-√{[x+(1/x)]^2-4}

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已知0<x<1,化简√{[x-(1/x)]^2+4}-√{[x+(1/x)]^2-4}已知0<x<1,化简√{[x-(1/x)]^2+4}-√{[x+(1/x)]^2-4}已知0<x<1,化简√{[x

已知0<x<1,化简√{[x-(1/x)]^2+4}-√{[x+(1/x)]^2-4}
已知0<x<1,化简√{[x-(1/x)]^2+4}-√{[x+(1/x)]^2-4}

已知0<x<1,化简√{[x-(1/x)]^2+4}-√{[x+(1/x)]^2-4}
[x-(1/x)]^2+4=x²-2+1/x²+4=x²+2+1/x²=(x+1/x)²
[x+(1/x)]^2-4=x²+2+1/x²-4=x²-2+1/x²=(x-1/x)²
∵ 0

√{[x-(1/x)]^2+4}-√{[x+(1/x)]^2-4}
=√{[x+(1/x)]^2}-√{[x-(1/x)]^2}
=x-1/x-(1/x-x)
=2x

【参考答案】

∵当0∴原式=√[x²-2+(1/x²)+4]-√[x²+2+(1/x²)-4]
=√[x+(1/x)]²-√[x-(1/x)]²
=x+(1/x)-[(1/x)-x]
=x+(1/x)-(1/x)+x
=2x