Question

已知a+b+2(1-a-b)=3(1-b-a)-4(b-1+a),求代数式36(a+b)2-12(a+b)+1的值

Answer 1

a+b+2(1-a-b)=3(1-b-a)-4(b-1+a)

a+b+2-2(a+b)=3-3(a+b)-4[(a+b)-1]

a+b+2-2(a+b)=3-3(a+b)-4(a+b)+4

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

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

6(a+b)=5

a+b=5/6

36(a+b)^2-12(a+b)+1

=[6(a+b)-1]^2

=[6*5/6-1]^2

=4^2

=16