Question

[1+(a-b)分之2b]^2乘以[1-(a+b分之2b)]^2

Answer 1

[1+2b/(a-b)]^2×[1-2b/(a+b)]^2

=[(a-b)/(a-b)+2b/(a-b)]^2×[(a+b)/(a+b)-2b/(a+b)]^2

=[(a+b)/(a-b)]^2×[(a-b)/(a+b)]^2

=1