Hằng đẳng thức
(a+b+c)^2= a^2+ b^2+ c^2 + 2ab +2ac +2bc
(a-b-c)^2= a^2+ b^2+ c^2 -2ab - 2ac +2bc
(a+b-c)^2=a^2+ b^2+ c^2 +2ab - 2ac -2bc

2x - x^2 - 2
<=> - x^2 + 2x - 1 - 1
<=> -(x^2 - 2x + 1) - 1
<=> -(x - 1)^2 - 1 < 0 (Do -(x - 1)^2 ≤ 0 ∀ x)
Vậy 2x - x^2 - 2 < 0 ∀ x

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac
(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ac
(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ac
(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ac

nhiều lắm bạn à
(1) (a+b+c)2=a2+b2+c2+2ab+2bc+2ac
(2) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) (a−b−c)2=a2+b2+c2−2ab−2ac+2bc
(4) a3+b3=(a+b)3−3ab(a+b)
(5) a3−b3=(a−b)3+3ab(a−b)
(6) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)

(8) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2
(10) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab2+bc2+ca2−a2b−b2c−c2a=[(a−b)3+(b−c)3+(c−a)3]/3
(13) an−bn=(a−b)(an−1+an−2b+an−3b2+...+a2bn−3+abn−2+bn−1)
(12)ab3+bc3+ca3−a3b−b3c−c3a={(a+b+c)[(a−b)3+(b−c)3+(c−a)3]}/3
(14) Với n lẻ:
an+bn=(a+b)(an−1−an−2b+an−3b2−...+a2bn−3−abn−2+bn−1)