化简求值：(2a-b+1)(2a-b-1)-(a+2b)(a-b),其中a,b满足la+b-3l+(ab+2)^2=0

即a+b-3=0
ab+2=0
a+b=3
ab=-2
原式=(2a-b)²-1²-(a²-ab+2ab-2b²)
=4a²-4ab+b²-1-a²-ab+2b²
=3a²+3b²-5ab-1
=3(a²+b²+2ab)-11ab-1
=3(a+b)²-11ab-1
=27+22-1
=48

因为：la+b-3l+(ab+2)^2=0，且la+b-3l≥0，(ab+2)^2≥0，所以：la+b-3l=0，(ab+2)^2=0a+b-3=0，ab+2=0即：a+b=3，ab=-2 (2a-b+1)(2a-b-1)-(a+2b)(a-b)=[（2a-b)??-1]-（a??-ab+2ab-2b??)
=4a??+b??-4ab-1-a??-ab+2b??
=3a??+3b??-5ab-1=3（a??+b??）-5ab-1=3[(a+b)??-2ab）]-5ab-1=3(a+b)??-11ab-1=3*3??-11*(-2)-1=50