原式={(3x+4)/[(x+1)(x-1)]-2(x+1)/[(x+1)(x-1)]}/[(x+2)/(x-1)²]
={(3x+4-2x-2}/[(x+1)(x-1)]}/[(x+2)/(x-1)²]
={(x+2}/[(x+1)(x-1)]}/[(x+2)/(x-1)²]
=(x-1)/(x+1)
把x=根号2代入得
原式=(根号2-1)/(根号2+1)
=(根号2-1)²/[(根号2+1)(根号2-1}]
=(根号2-1)²
=3-2根号2
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