(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=
问题描述:
(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
(a+b)(a+b+1)+1/4-y²
=(a+b+y+1/2)(a+b-y+1/2)
a²-2ab+b²-2a+2b+1
=(a-b-1)^2
(2^102-2^101)/(2^98-2^99)
=-8
800²-1600*798+798²
=4
答
1 =(a+b)²+(a+b)+1/4+y²
=(a+b+1/2)²-y²
=(a+b+1/2+y)(a+b+1/2-y)
2 =(a-b)²-2(a-b)+1
=(a-b-1)²
3 =2^101/(-2^98)
=-2^3
-8
4 =800²-2*800*798+798²
=(800-798)²
=2²
=4
答
(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