First simplify, then evaluate: a - (2-A) - (a + 1) (A-1) + (A-1) 2, where a = root 3

First simplify, then evaluate: a - (2-A) - (a + 1) (A-1) + (A-1) 2, where a = root 3

The original formula = a - (2-A) - (A2-1) + (a2-2a + 1) = A-2 + A-A2 + 1 + a2-2a + 1 = 0

Let AB satisfy AB = 1, and prove A2 + B2 ≥ a + B

A & # 178; + B & # 178; - (a + b) = A & # 178; + B & # 178; + 2Ab - (a + b) - 2Ab = (a + b) &# 178; - (a + b) - 2 = (a + b-2) (a + B + 1) a, B are positive, a + B ≥ 2 √ (AB) = 2, so a + B-2 ≥ 0A + B + 1 > 0 (a + b-2) (a + B + 1) ≥ 0A & # 178; + B & # 178; ≥ a + B

A2 + B2 = 4, ab = 1, then (a + b) 2 =?

4

Calculation: (a-b) (A2 + AB + B2)

The original formula is A3 + A2B + ab2-a2b-ab2-b3 = a3-b3