Question

已知:(a+2)的平方+|a+b+5|=0,求3a的平方b-[2a的平方b-(2ab-a的平方b)-4a的平方]-ab的值

Answer 1

a+2=0,a+b+5=0
a=-2,b=-3

3a的平方b-[2a的平方b-(2ab-a的平方b)-4a的平方]-ab
=3a^2b-2a^2b+2ab-a^2b+4a^2-ab
=4a^2+ab
=4*(-2)^2+(-2)*(-3)
=16+6
=22

Answer 2

(a+2)的平方+|a+b+5|=0
a+2=0,a+b+5=0
a=-2
b=-5-a=-3

3a的平方b-[2a的平方b-(2ab-a的平方b)-4a的平方]-ab
=3a^2b-2a^b+2ab-a^2b+4a^2-ab
=4a^2+ab
=4*(-2)^2+(-2)*(-3)
=16+6
=22

Answer 3

(a+2)的平方+|a+b+5|=0
(a+2)^2=0,
a+2=0,a=-2
|a+b+5|=0,
a+b+5=0,b=-3

3a的平方b-[2a的平方b-(2ab-a的平方b)-4a的平方]-ab
=3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab
=3a^2b-2a^2b+2ab-a^2b+4a^2-ab
=4a^2+ab
代入a=-2,b=-3
原式=4(-2)^2+(-2)(-3)
=16+6=22

Answer 4

∵(a+2)^2+|a+b+5|=0
∴(a+2)^2=0,|a+b+5|=0
∴a=-2,b=-3
∴3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab
=3a^2b-2a^2b-2ab-a^2b+4a^2-ab
=4a^2+ab
=4*(-2)^2+(-2)*(-3)
=22