4x的平方+kx-5=(x+1)·A,则A=

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4x的平方+kx-5=(x+1)·A,则A=

4x的平方+kx-5=(x+1)·A,
则A=4x-5

设A=ax+b即4x²+kx-5=ax²+﹙a+b)x+b,对应项系数相等,即a=4,b=-5,k=a+b=-1,∴A =4x-5

==>4xx-4+kx-1=A(x+1)
==>4(x+1)(x-1)+kx-1=A(x+1)
==>k=-1, (x+1)(4x-4)-x-1=A(x+1)
==>(x+1)(4x-5)=A(x+1)
==>A=4x-5