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(1)已知a,b满足a2+b2+4a-8b+20=0,试分解(x2+y2)-(b+axy);
(2)计算:(1-
1
22
)(1-
1
32
)(1-
1
42
)…(1-
1
20082
)(1-
1
20092
);
(3)设a=1999x+1998,b=1999x+1999,c=1999x+2000,求a2+b2+c2-ab-ac-bc的值.
人气:465 ℃ 时间:2020-04-03 18:31:17
解答
(1)a2+b2+4a-8b+20=0,(a+2)2+(b-4)2=0,所以a=-2,b=4,(x2+y2)-(4-2xy)=x2+y2+2xy-4=(x+y)2-4=(x+y+2)(x+y-2);(2)原式=(1-12)×(1+12)×(1-13)×(1+13)×(1-14)×(1+14)×…×(1-...
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