已知a²;+b²;=12,ab=-3,則(a+b)²;=?,(a-b)²;=?

已知a²;+b²;=12,ab=-3,則(a+b)²;=?,(a-b)²;=?


(a+b)²;=a^2+b^2+2ab=12-3*2=6
(a-b)²;=a^2+b^2-2ab=12-2*(-3)=18



a-1/a+2÷a²;-2a+1/a²;-4×a+3/a²;+a-6


a-1/a+2÷a²;-2a+1/a²;-4×a+3/a²;+a-6
=a-1/a+2÷(a-1)²;/(a+2)(a-2)×(a+3)/(a-2)(a+3)
=(a-1)/(a+2)×(a+2)(a-2)/(a-1)²;×1/(a-2)
=(a-2)/(a-1)×1/(a-2)
=1/(a-1)



(2a-1)²;-2×(2a-1)×(a-6)+(a-6)²;


(2a-1)²;-2×(2a-1)×(a-6)+(a-6)²;
=(2a-1-a-6)²;
=(a-7)²;