整式的加减数学题,

1:已知A=x的三次方-5x的平方 B=x的平方-11x+6,

求,(1)A+2B, (2)当x= -1 时, A+5B的值.

2:已知xy+x= -1 ,xy-y=2 求代数式,-x-[2y-2(xy+x)的平方+3x]+2[x+(xy-y)的平方]的值.

（过程简要）

3：化简，5a-{-36+[bc-2a-(a-c)}-[9a-(7a+c)]

(_过程简要)

我急用阿,各位谢谢了..

1:

A=x³-5x²,B=x²-11x+6

A+2B

=x³-5x²+2(x²-11x+6)

=x³-5x²+2x²-22x+12

=x³-3x²-22x+12

(2)

A+5B
=x³-5x²+5(x²-11x+6)

=x³-55x+30

当x=-1时

=(-1)³-55×(-1)+30

=-1+55+30

=84

2:

因为yx+x=-1,xy-y=2

所以(yx+x)-(xy-y)=x+y=-1-2=-3

-x-[2y-2(xy+x)²+3x]+2[x+(xy-x)²]

代入yx+x=-1,xy-y=2

=-x-[2y-2+3x]+2[x+4]

=-x-2y+2-3x+2x+4

=-2x-2y+6

=-2(x+y）+6

代入x+y=-2

=-2×(-3)+6

=12

(3)

5a-{-36+[bc-2a-(a-c)}-[9a-(7a+c)]

=5a-{-36+[bc-3a+c)}-[2a-c)]

=5a-{-36+[bc-3a+c-2a+c)]

=5a-[-36+(bc-5a+2c)]

=5a+36-bc+5a-2c

=10a-bc-2c+36