若(x2+px+28/3)(x2−3x+q)=0的积中不含x2与x3项, (1)求p、q的值; (2)求代数式(-2p2q)3+(3pq)-1+p2010q2012的值.

问题描述:

(x2+px+

28
3
)(x2−3x+q)=0的积中不含x2与x3项,
(1)求p、q的值;
(2)求代数式(-2p2q)3+(3pq)-1+p2010q2012的值.

(1)(x2+px+

28
3
)(x2−3x+q)=0,
x4-3x3+qx2+px3-3px2+pqx+
28
3
x2-28x+
28
3
q=0,
x4+(p-3)x3+(q-3p+
28
3
)x2+(pq-28)x+
28
3
q=0,
因为它的积中不含有x2与x3项,
则有,p-3=0,q-3p+
28
3
=0
解得,p=3,q=-
1
3

(2)(-2p2q)3+(3pq)-1+p2010q2012
=[-2×9×(-
1
3
)]3+[3×3×(-
1
3
)]-1+(pq)2010q2
=63-
1
3
+(-
1
3
×3)2010•(-
1
3
2
=216-
1
3
+1×
1
9

=216-
1
3
+
1
9

=215
7
9