Find the value of (a 2+2ab+2a) (b 2+2ab+2b).

Derived from the fact that A and B are two real numbers of the equation x 2+2x-5 = 0:

AB=-5，A+B=-2

A^2+2AB+2A)(B^2+2AB+2B)

=AB(A+2B+2)(B+2A+2)

=-5(-2+B+2)(-2+A+2)

=-5AB

=25

2, 1/2(x+y+z) squared+1/2 (x-y-z) (x-y+z)-z (x+y), where x-y=6 and xy=2 1. Be specific.

Simplify:

1/2(x+y+z) squared+1/2 (x-y-z) (x-y+z)-z (x+y) =1

1/2[(x+y) square +2z(x+y)+z square ]+ 1/2[(x-y) square -z square ]-z(x+y)=]

1/2(x+y) square+1/2(x-y) square =x square +y square.

From x-y=6, xy=2 1, x+y = (x-y)+2xy = 78.

3. Find the value of 2ab-2a 2-4b 2-7 from a 2-ab+2b 2 = 3.

2ab-2a^2-4b^2-7

=2(ab-a^2-2b^2)-7

=-2(a^2-ab+2b^2)-7

=(-2)*3-7

=-6-7=- 13

4. If A = 2x 2+3xy-2x-3, B =-x 2+xy+2, and the value of 3A+6B has nothing to do with X, find the value of y..

Solution:

3a+6b=6x^2+9xy-6x-9-6x^2+6xy+ 12

= 15xy-6x+3

=x( 15y-6)+3

5.9x+6x 2-3 (x-2/3x 2), where x=-2.

9x+6x？ 0? 5 -3(x-2/3x？ 0? 5)

=9x+6x？ 0? 5-3x+2x？ 0? five

=8x？ 0? 5+6 times

=8×(-2)? 0? 5+6×(-2)

=32- 12

=20

6.1/4 (-4x2+2x-8)-(1/2x-1), where x= 1/2.

1/4(-4x？ 0? 5+2x-8)-( 1/2x- 1)

=-x？ 0? 5+ 1/2x-2- 1/2x+ 1

=-x？ 0? 5- 1

=-( 1/2)? 0? 5- 1

=- 1/4- 1

=-5/4

7.3x' y-[2x' y-(2xyz-x' z)-4x' z]-XYZ, where x =-2, y =-3, z = 1,

:3x ' y-[2x ' y-(2xyz-x ' z)-4x ' z]-XYZ

=3x'y-2x'y+2xyz-x'z+4x'z-xyz

=x'y-xyz+3x'z

=4*(-3)-2*3* 1+3*4* 1

=- 12-6+ 12

=-6

8.(5a 2-3b 2)+(a 2+b 2)-(5a 2+3b 2), where a =- 1 and b = 1.

=5a^2-3b^2+a^2+b^2-5a^2-3b^2

=a^2-5b^2

=(- 1)^2-5* 1^2

= 1-5

=-4

9,2 (A2B+AB 2)-2 (A2B-1)-2AB 2-2, where a =-2 and b = 2.

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

=0

10 squared, (x-21y-1) (x-21y+1)-(x-21y-1).

Where x = 1.7 and y = 3.9 (simplify first and then evaluate).

[(X-2 is 1Y)- 1][(X+2 is 1Y)+ 1]-(X-2 is1y-0/) square.

= (x+ 1Y) square-1-(x- 1Y) square +2 (x- 1Y)- 1.

= (x+ 1Y) square -(x- 1Y) square +2 (x- 1y) -(X-2

=2XY+2X-Y-2

=3.9*2.4+ 1.4

= 10.76