化简求值;{[(a-2)⼀(a^2+2a)]-[(a-1)⼀(a^2+4a+4)]}除以(a-4)⼀(a+2),其中a满足a^2+2a-1=0

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2024-12-01 10:26:37
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回答1:

{[(a-2)/(a^2+2a)]-[(a-1)/(a^2+4a+4)]}除以(a-4)/(a+2)
={[(a-2)/a(a+2)]-[(a-1)/(a+2)²]}除以(a-4)/(a+2)
=[(a+2)(a-2)-a(a-1)]/a(a+2)²×(a+2)/(a-4)
=1/(a²+2a)
=1