(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
问题描述:
(1)已知a,b满足a2+b2+4a-8b+20=0,试分解(x2+y2)-(b+axy);
(2)计算:(1-
)(1-1 22
)(1-1 32
)…(1-1 42
)(1-1 20082
);1 20092
(3)设a=1999x+1998,b=1999x+1999,c=1999x+2000,求a2+b2+c2-ab-ac-bc的值.
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-
)×(1+1 2
)×(1-1 2
)×(1+1 3
)×(1-1 3
)×(1+1 4
)×…×(1-1 4
)×(1+1 2008
)×(1-1 2008
)×(1+1 2009
)1 2009
=
×1 2
×3 2
×2 3
×4 3
×…×3 4
×2007 2008
×2009 2008
×2008 2009
2010 2009
=
×1 2
2010 2009
=
;1005 2009
(3)2(a2+b2+c2-ab-ac-bc)
=(a-b)2+(a-c)2+(b-c)2;
当a=1999x+1998,b=1999x+1999,c=1999x+2000时,
(a-b)2+(a-c)2+(b-c)2
=(-1)2+(-2)2+(-1)2
=1+4+1
=6.
所以a2+b2+c2-ab-ac-bc=6÷2=3.