解:(1)∵(x+2)2+|y+1|=0, ∴x+2=0,y+1=0, ∴x=﹣2,y=﹣1, 5xy2﹣{2x2y﹣[3xy2﹣(4xy2﹣2x2y)]}, =5xy2﹣[2x2y﹣(3xy2﹣4xy2+2x2y)], =5xy2﹣(2x2y﹣3xy2+4xy2﹣2x2y), =5xy2﹣2x2y+3xy2﹣4xy2+2x2y, =4xy2, 把x=﹣2,y=﹣1,代入上式, 原式=4×(﹣2)×(﹣1)2=﹣8; (2)观察发现:连续整数的和等于第一项与最后一项的和与最后一项的倍数除以2, ∴1+2+3+…+n=, 1+2+3+…+1000=(1+1000)×1000 ÷2=500500.
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