Question

已知(a-b+1)的平方=-4a的平方+4ab-b的平方,则〔(a+3b)(3b-a)-(2a-3b)的平方〕/5a=_____

Answer 1

(a-b+1)^2=-4a^2+4ab-b^2=-(4a^2-4ab+b^2)=-(2a-b)^2
所以有
(a-b+1)^2+(2a-b)^2=0
a-b+1=0 2a-b=0
a-b+1=0 b=2a代入前式
a-2a+1=0
a=1
b=2

〔(a+3b)(3b-a)-(2a-3b)的平方〕/5a
=(3ab-a^2+9b^2-3ab-4a^2+12ab-9b^2)/5a
=(-5a^2+12ab)/5a
=-a+12b/5
=-1+12*2/5
=19/5

Answer 2

(a-b+1)^2=-4a^2+4ab-b^2=-(4a^2-4ab+b^2)=-(2a-b)^2
(a-b+1)^2+(2a-b)^2=0
a-b+1=0 2a-b=0
a-b+1=0 b=2a
a-2a+1=0
a=1
b=2
(a+3b)(3b-a)-(2a-3b)的平方〕/5a=-(a+3b)(a-3b)/5-(2a-3b)^2/5=-(1-36)/5-16/5=19/5