
分式的乘除
1个回答
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(1)原式=x分之(x-1) × [x(x-1)]分之1
=x²分之1
.
(2)原式=(x+1)²分之[(x-1)+2]*[(x-1)-2] × [(x+3)(x-3)]分之[x(x+1)]
=(x+1)分之(x+1)(x-3) × [(x+3)(x-3)]分之x
=(x+3)分之x
=x²分之1
.
(2)原式=(x+1)²分之[(x-1)+2]*[(x-1)-2] × [(x+3)(x-3)]分之[x(x+1)]
=(x+1)分之(x+1)(x-3) × [(x+3)(x-3)]分之x
=(x+3)分之x
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