原式=√[(x²-2+1/x²+4)]-√[(x²+2+1/x²-4)]
=√(x²+2+1/x²)-√(x²-2+1/x²)
=√(x+1/x)²-√(x-1/x)²
=|x+1/x|+|x-1/x|
由两个绝对值得符号,有四种情况
原式=x+1/x+x-1/x=2x
或原式=x+1/x-x+1/x=2/x
或原式=-x-1/x+x-1/x=-2/x
或原式=-x-1/x-x+1/x=-2x
原式=√[(x²-2+1/x²+4)]-√[(x²+2+1/x²-4)]
=√(x²+2+1/x²)-√(x²-2+1/x²)
=√(x+1/x)²-√(x-1/x)²
=|x+1/x|+|x-1/x|
由两个绝对值得符号,有四种情况
原式=x+1/x+x-1/x=2x
或原式=x+1/x-x+1/x=2/x
或原式=-x-1/x+x-1/x=-2/x
或原式=-x-1/x-x+1/x=-2x