Question

当a=1,b=2,求多项式……

谢谢啦~当a=1,b=2,求多项式5ab—二分之九a³b²－四分之九ab+二分之一a³b²—四分之十一ab—a³b—5的值。

Answer 1

5ab—二分之九a³b²－四分之九ab+二分之一a³b²—四分之十一ab—a³b—5
＝5x1x2-(9/2)x4-(9/4)x2+(1/2)x4-(11/4)x2-2-5
=10-18-9/2+2-11/2-2-5
=(10-18+2-2-5)-(9/2+11/2)
=-13-10
=-23

Answer 2

-23

Answer 3

答案是-51

Answer 4

5ab—9a³b²/2－9ab/4+a³b²/2—11ab/4—a³b—5
=(5ab－9ab/4—11ab/4)+(—9a³b²/2+a³b²/2)—a³b—5
=0-4a³b²-a³b-5
=-4*1³*2²-1³*2-5
=-16-2-5
=-23