∵x^2+4x-5=0
∴x^2+4x=5
(2x+3)(2x-1)-2(x-1)²+1
=4x^2+6x-2x-3-2(x^2-2x+1)+1
=4x^2+4x-3-2x^2+4x-2+1
=2x^2+8x-4
=2(x^2+4x-2)
=2(5-2)
=2*3
=6
x²+4x-5=0,求(2x+3)(2x-1)-2(x-1)²+1的值
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