题目内容
(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-
)…(1-
)(1-
);
(3)设a=1999x+1998,b=1999x+1999,c=1999x+2000,求a2+b2+c2-ab-ac-bc的值.
(2)计算:(1-
1 |
22 |
1 |
32 |
1 |
42 |
1 |
20082 |
1 |
20092 |
(3)设a=1999x+1998,b=1999x+1999,c=1999x+2000,求a2+b2+c2-ab-ac-bc的值.
(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-
)×(1+
)×(1-
)×(1+
)×…×(1-
)×(1+
)×(1-
)×(1+
)
=
×
×
×
×
×…×
×
×
×
=
×
=
;
(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.
(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 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
4 |
1 |
2008 |
1 |
2008 |
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.
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