
计算:[(x√y-y√x)/x√y+y√x]-[(y√x+x√y)/y√x-x√y]
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解:原式=(x√y-y√x)/(x√y+y√x)-(y√x+x√y)/(y√x-x√y) 分母有理化
=(x√y-y√x)²/[(x√y+y√x)(x√y-y√x)]-(y√x+x√y)²/[(y√x+x√y)(y√x-x√y)]
=[x²y-2xy√(xy)+xy²]/(x²y-xy²)-[xy²+2xy√(xy)+x²y]/(xy²-x²y)
=[x²y-2xy√(xy)+xy²]/(x²y-xy²)+[xy²+2xy√(xy)+x²y]/(x²y-xy²)
=[x²y-2xy√(xy)+xy²+xy²+2xy√(xy)+x²y]/(x²y-xy²)
=(2x²y+2xy²)/(x²y-xy²)
=[2xy(x+y)]/[xy(x-y)]
=2(x+y)/(x-y)
=(x√y-y√x)²/[(x√y+y√x)(x√y-y√x)]-(y√x+x√y)²/[(y√x+x√y)(y√x-x√y)]
=[x²y-2xy√(xy)+xy²]/(x²y-xy²)-[xy²+2xy√(xy)+x²y]/(xy²-x²y)
=[x²y-2xy√(xy)+xy²]/(x²y-xy²)+[xy²+2xy√(xy)+x²y]/(x²y-xy²)
=[x²y-2xy√(xy)+xy²+xy²+2xy√(xy)+x²y]/(x²y-xy²)
=(2x²y+2xy²)/(x²y-xy²)
=[2xy(x+y)]/[xy(x-y)]
=2(x+y)/(x-y)
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