(1)原式=(1/x+x+1)(1/√x+√x)
=(1+x+x²)/x·(1-x)/√x
=(1-x³)/x√x
(2)设xˆ-2/3=a yˆ-2/3=b
原式=(a³+b³)/(a+b)-(a³-b³)/(a-b)
=a²-ab+b²-(a²+ab+b²)
=-2ab
=-2xˆ-2/3·y-2/3
(1)原式=(1/x+x+1)(1/√x+√x)
=(1+x+x²)/x·(1-x)/√x
=(1-x³)/x√x
(2)设xˆ-2/3=a yˆ-2/3=b
原式=(a³+b³)/(a+b)-(a³-b³)/(a-b)
=a²-ab+b²-(a²+ab+b²)
=-2ab
=-2xˆ-2/3·y-2/3