Let a, B, C be the trilateral length of △ ABC, satisfying a & sup2; + 2B & sup2; + C & sup2; - 2b (a + C) = 0

Let a, B, C be the trilateral length of △ ABC, satisfying a & sup2; + 2B & sup2; + C & sup2; - 2b (a + C) = 0

a²+b²+b²+c²-2ab-2bc=0
(a²-2ab+b²)+(b²-2bc+c²)=0
(a-b)²+(b-c)²=0
The square is greater than or equal to 0, and the sum is equal to 0
If one is greater than 0, then the other is less than 0
So both are equal to zero
So A-B = 0, B-C = 0
a=b,b=c
Then a = b = C
So it's an equilateral triangle

Given that a, B and C are the lengths of the three sides of △ ABC and satisfy A2 + 2B2 + c2-2b (a + C) = 0, then the shape of the triangle is______ ．

From the known condition A2 + 2B2 + c2-2b (a + C) = 0, it is concluded that, (a-b) 2 + (B-C) 2 = 0  A-B = 0, B-C = 0, that is, a = B, B = C  a = b = C, so the answer is equilateral triangle

If a & sup2; + 2B & sup2; - 2ab-2bc + C & sup2; = 0, try to judge the shape of triangle (a, B, C are the three sides of △ ABC respectively), and explain the reason

It is an equilateral triangle. A & sup2; + 2B & sup2; - 2ab-2bc + C & sup2; = 0 is written as a & sup2; + B & sup2; + B & sup2; - 2ab-2bc + C & sup2; = 0. This equation is two complete square expressions. By simplifying, we can get (a-b) & sup2; + (B-C) & sup2; = 0. The square of a number can't be less than 0, it can only be greater than or

Calculation or simplification: [(2a + 2) / (A-1)] / (a + 1) - [(A & sup2; - 1) / (A & sup2; - 2A + 1)]

The original formula = [2 (a + 1) / (A-1) × [1 / (a + 1)] - (a + 1) (A-1) / (A-1) & sup2;
=2/(a-1)-(a+1)/(a-1)
=(2-a-1)/(a-1)
=-(a-1)/(a-1)
=-1