Question

第5题，怎么做，数学课题

Answer 1

详细解答，请见下图：

Answer 2

[a+1 - (4a-5)/(a-1) ] /[ 1/(a-1) - 2/(a^2-a) ]
=[ (a^2-4a+4)/(a-1) ] /[ (a-2a)/(a^2-a) ]
=[ (a-2)^2/(a-1) ] /{ [a(a-2)]/[(a(a-1) ] }
=[ (a-2)^2/(a-1) ] / [(a-2)/(a-1) ]
=[ (a-2)^2/(a-1) ] . [(a-1)/(a-2) ]
=a-2

Answer 3

5、原式
=[(a+1)(a-1)/(a-1) -(4a-5)/(a-1)] ÷[a/a(a-1) -2/a(a-1)]
=[(a²-1-4a+5)/(a-1)] ÷[(a-2)/a(a-1)]
=[(a-2)²/(a-1)]•[a(a-1)/(a-2)]
=a(a-2)
=a²-2a
=(-1)²-2×(-1)
=1+2
=3

Answer 4

原式={【（a+1)(a-1）-（4a-5)】/a-1}÷(a-2)/a(a-1)
=【(a²-1-4a+5)/a-1】÷(a-2)/a(a-1)
=（a²-4a+4）/a-1÷(a-2)/a(a-1)
=（a-2）²/a-1×a(a-1)/(a-2)
=a（a-2）
=a²-2a
当a=-1时
原式=（-1）²-2×（-1）=1+2=3
采纳呗

Answer 5