(a^2+2a+ 1)+(b^2-4b+4)=0

(a+ 1)^2+(b-2)^2=0

Square is greater than or equal to 0, and sum is equal to 0. If one is greater than 0 and the other is less than 0, it is not valid.

So both are equal to 0.

So a+ 1 = 0 and b-2 = 0.

a=- 1，b=2

b^a=2^(- 1)= 1/2

2.( 16x^2y^5-2xy^3+ 1/4xy)/(- 1/4xy)

=-4xy*( 16x^2y^5-2xy^3+ 1/4xy)

=-64x^3y^6+8x^2y^4- 1

3. Fill in the blank: _ 12x 4- 16x 3+8x 2 _ _ divided by (-4x 2) =-3x 2+4x-2.

4. Let these three numbers be n- 1, n, n+ 1.

Then (n-1) 3+n 3+(n+1) 3.

=n^3-3n^2+3n- 1+n^3+n^3+3n^2+3n+ 1

=3n^3+6n

=3n(n^2+2)

N is the middle

So it can be divisible by three times the middle integer.

5.(4a^2-6ab+9b^2)*(-0.5a)

=4a^2*(-0.5a)-6ab *(0.5a)+9b^2*(-0.5a)

=-2a^3+3a^2b-4.5ab^2