Question

已知（a+1）的平方—（3a的平方+4ab+4b的平方+2）=0，求a+b的值

Answer 1

(a+1)^2-(3a^2+4ab+4b^2+2)=0
a^2+2a+1-3a^2-4ab-4b^2-2=0
2a^2-2a+1+4ab+4b^2=0
(a^2-2a+1)+(a^2+4ab+4b^2)=0
(a-1)^2+(a+2b)^2=0
a-1=0 a+2b=0 联立
解得a=1 b=-1/2
a+b=1-1/2=1/2

Answer 2

(a+1)^2-(3a^2+4ab+4b^2+2)
=(a^2+2a+1)-(3a^2+4ab+4b^2+2)
=-2a^2-4ab-4b^2+2a-1
=-(a^2-2a+1)-(a^2+4ab+4b^2)
=-(a-1)^2-(a+2b)^2=0
∴a-1=a+2b=0, a=1,b=-1/2
∴a+b=1-1/2=1/2

Answer 3

(a+1)�0�5-(3a�0�5+4ab+4b�0�5+2)=0,
2a�0�5+4ab+4b�0�5+1-2a=0,
(a-2b)�0�5+(a-1)�0�5=0,
a=1,b=1/2,
a+b=3/2.

Answer 4

a^2+2a+1-3a^2-4ab-4b^2-2=0
2a^2-2a+1+4ab+4b^2=0
(a+2b)^2+(a-1)^2=0
a=1 b=-1\2
a+b=1\2