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:
若(x2+px+
)(x2−3x+q)=0的积中不含x2与x3项,28 3
(1)求p、q的值;
(2)求代数式(-2p2q)3+(3pq)-1+p2010q2012的值.
答
(1)(x2+px+
)(x2−3x+q)=0,28 3
x4-3x3+qx2+px3-3px2+pqx+
x2-28x+28 3
q=0,28 3
x4+(p-3)x3+(q-3p+
)x2+(pq-28)x+28 3
q=0,28 3
因为它的积中不含有x2与x3项,
则有,p-3=0,q-3p+
=028 3
解得,p=3,q=-
,1 3
(2)(-2p2q)3+(3pq)-1+p2010q2012
=[-2×9×(-
)]3+[3×3×(-1 3
)]-1+(pq)2010q21 3
=63-
+(-1 3
×3)2010•(-1 3
)21 3
=216-
+1×1 3
1 9
=216-
+1 3
1 9
=215
.7 9