因式分解:[x(x²y²-xy)-y(x²-x³y)]÷3x²y
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原式=[x²y(xy-1)-x²y(1-xy)]÷3x²y
=(xy-1-1+xy)÷3
=(2xy+2)/3
=(xy-1-1+xy)÷3
=(2xy+2)/3
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[x(x²y²-xy)-y(x²-x³y)]÷3x²y
=[x³y²-x²y-yx²+x³y²]÷3x²y
=[2x³y²-2x²y]÷3x²y
=【2xy-2】/3
=[x³y²-x²y-yx²+x³y²]÷3x²y
=[2x³y²-2x²y]÷3x²y
=【2xy-2】/3
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:[x(x²y²-xy)-y(x²-x³y)]÷3x²y
=(x³y²-x²y-x²y+x³y²)÷3x²y
=(2x³y²-2x²y)÷3x²y
=2x²y(xy-1)÷3x²y
=2/3(xy-1)
=(x³y²-x²y-x²y+x³y²)÷3x²y
=(2x³y²-2x²y)÷3x²y
=2x²y(xy-1)÷3x²y
=2/3(xy-1)
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[x(x²y²-xy)-y(x²-x³y)]÷3x²y
=[x²y(xy-1)-x²y(1-xy)]÷3x²y
=x²y[(xy-1)-(1-xy)]÷3x²y
=x²y[2(xy-1)]÷3x²y
=2(xy-1)/3
=[x²y(xy-1)-x²y(1-xy)]÷3x²y
=x²y[(xy-1)-(1-xy)]÷3x²y
=x²y[2(xy-1)]÷3x²y
=2(xy-1)/3
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