将原式展开得:x4+(p-3)x^3+(q-3p+28/3)x^2+pqx-28x+28/3x
因为积中不含x的二次方与x的三次方,
令p-3=0
q-3p+28/3=0
即:
p=3
q=-1/3
(-2p²q)³+(3pq)负一次方+P的2011次方q的2012次方
=[2*9*(1/3)]3+[3*3*(-1/3)]-1+[3*(-1/3)]2011*(-1/3)
=8*33-1/3-1*(-1/3)
=297
将原式展开得:x4+(p-3)x^3+(q-3p+28/3)x^2+pqx-28x+28/3x
因为积中不含x的二次方与x的三次方,
令p-3=0
q-3p+28/3=0
即:
p=3
q=-1/3
(-2p²q)³+(3pq)负一次方+P的2011次方q的2012次方
=[2*9*(1/3)]3+[3*3*(-1/3)]-1+[3*(-1/3)]2011*(-1/3)
=8*33-1/3-1*(-1/3)
=297