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分时化简： [1⼀（2x+2）]-[x⼀（x²-1）]+[1⼀（x-1）]

2026年09月24日 04:49

有2个网友回答

网友（1）：

[1/（2x+2）]-[x/（x2-1）]+[1/（x-1）]
=[1/2 * （x+1）]-[x/（x-1）(x+1)]+(x+1)/[(x+1)(x-1)]
=[1/2 * （x+1）]+1/[(x+1)(x-1)]
=(x-1)/[2(x-1)(x+1)]+2/[2(x+1)(x-1)]
=(x+1)/[2(x-1)(x+1)]
=1/[2(x-1)]

网友（2）：

[1/（2x+2）]-[x/（x²-1）]+[1/（x-1）]
=(x-1)/2(x+1)(x-1)-2x/2(x+1)(x-1)+2(x+1)/2(x+1)(x-1)
=(x-1-2x+2x+2)/2(x+1)(x-1)
=1/2(x-1)

分时化简： [1⼀（2x+2）]-[x⼀（x&sup2;-1）]+[1⼀（x-1）]

分时化简： [1⼀（2x+2）]-[x⼀（x²-1）]+[1⼀（x-1）]