（a+b)(a+b+1)+1/4-y²

题目

（a+b)(a+b+1)+1/4-y²
=
a²-2ab+b²-2a+2b+1
=
(2^102-2^101)/(2^98-2^99)
=
800²-1600*798+798²
=

答案

(a+b)(a+b+1)+1/4-y²
原式=(a+b)^2+(a+b)+1/4-y^2
=(a+b+1/2)^2-y^2
=(a+b+y+1/2)(a+b-y+1/2)
a²-2ab+b²-2a+2b+1
原式=(a-b)^2-2(a-b)+1
=(a-b-1)^2
(2^102-2^101)/(2^98-2^99)
原式=2^101(2-1)/2^98(1-2)
=-2^3
=-8
800²-1600*798+798²
原式=(800-798)^2
=2^2
=4