If C is an integer, try to judge the shape of △ ABC

If C is an integer, try to judge the shape of △ ABC

Radical A-1 + radical B-2 = 0
a－1=0,b－2=0
a=1,b=2
2－1＜c＜2＋1
If C is an integer
∴c=2
ABC is an isosceles triangle

It is known that in △ ABC, the opposite sides of ∠ a, B and C are a, B and C respectively, and A2 = B (B + C). (1) prove: ∠ a = 2 ∠ B; (2) if a = 3b, judge the shape of △ ABC

(1) It is proved that: A2 = B (B + C), that is BC2 = AC (AC + AB), extend CA to D, make ad = AB, connect dB. Then ∠ BAC = 2 ∠ d.. BC2 = AC · CD, BCAC = cdbc, and ∠ C = ∠ C, ｜ △ BCA ∽ DCB, so ∠ d = ∠ ABC. ｜ ∠ BAC = 2 ∠ ABC; (2) ∫ a = 3b, ｜ A2 = 3B 2, and A2 = B (b) ∫ a = 2

/The square of ab-2 / + (B-1) = 0, find the value of 1 of AB + (a + 1) (B + 1) 1 of AB + (a + 3) (B + 2) 1 of AB + (a + 2010) (B + 2010) 1 of ab

/The square of ab-2 / + (B-1) = 0, B-1 = 0, B = 1ab-2 = 0, a = 1 / 2Ab + (a + 1) (B + 1) 1 / 2Ab + (a + 2) (B + 2) 1 / 2Ab + +(a + 2010) (B + 2010) 1 / 2 = 1 / 1x2 + 1 / 2x3 + 1 / 3x4 + 1 / 4x5 + 1/2011x2012=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…… +1/2011-1/2012=1-1/...