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原式为[1/(a+1)+1/(a-1)]÷2a/(a²-2a+1)
=(a+1+a-1)/(a+1)(a-1)÷2a/(a-1)²
=2a/(a+1)(a-1) ×(a-1)²/2a
=a-1/a+1
a=1(应舍去,因为分母a-1=0)
取a=2
原式 =a-1/a+1
=2-1/2+1
=1/3
取a=3
原式 =a-1/a+1
=3-1/3+1
=1/2
=(a+1+a-1)/(a+1)(a-1)÷2a/(a-1)²
=2a/(a+1)(a-1) ×(a-1)²/2a
=a-1/a+1
a=1(应舍去,因为分母a-1=0)
取a=2
原式 =a-1/a+1
=2-1/2+1
=1/3
取a=3
原式 =a-1/a+1
=3-1/3+1
=1/2
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解:
(1/a+1 +1/a-1 )÷(2a/a²-2a+1)
=[(a-1+a+1)/(a+1)(a-1) ]÷[2a/(a-1)²]
=[2a/(a-1)(a+1)]×[(a-1)²/2a]
=a-1/a+1
当a=3时
原式=3-1/3+1
=1/2
(1/a+1 +1/a-1 )÷(2a/a²-2a+1)
=[(a-1+a+1)/(a+1)(a-1) ]÷[2a/(a-1)²]
=[2a/(a-1)(a+1)]×[(a-1)²/2a]
=a-1/a+1
当a=3时
原式=3-1/3+1
=1/2
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