It is known that a, B and C are the opposite sides of the three internal angles a, B and C of △ ABC, and a, B and C form an arithmetic sequence, B = 60 °, then the shape of △ ABC is______ ．

It is known that a, B and C are the opposite sides of the three internal angles a, B and C of △ ABC, and a, B and C form an arithmetic sequence, B = 60 °, then the shape of △ ABC is______ ．

∵ B = 60 °, ∵ a + C = 120 °. ∵ a, B, C are arithmetic sequence, ∵ 2B = a + C, from the sine theorem we can get 2sinb = 3 = Sina + sinc = 2sina + C2 & nbsp; Cosa − C2 = 3cosa − C2, ∵ Cosa − C2 = 1, and - 2 π 3 < a-c < 2 π 3, ∵ a-c = 0, so △ ABC is equilateral triangle, so the answer is regular triangle

Let m be a real number, f (x) = 2x ^ 2 + (x-m) | x-m |, H (x) = f (x) / x, X is not equal to 0, x x = 0 (1) if f (1) > = 4, find the value range of M (2)
When m > 0, we prove that h (x) is a monotone increasing function on [M, + ∞)

(1)
f(1)＝2＋(1－m)|1－m| ≥ 4
When m > 1, (1-m) (m-1) ≥ 2, there is no solution;
When m ≤ 1, (1-m) (1-m) ≥ 2, the solution is m ≤ 1 - √ 2
The value range of M is: m ≤ 1 - √ 2
(2)
∵m＞0,x ≥ m
∴h(x)＝f(x)/x
＝[2x²＋(x－m)(x－m)]/x
＝[2x²＋x²＋m²－2mx]/x
＝(3x²＋m²－2mx)/x
＝3x＋(m²/x)－2m
Let m ≤ x1 ≤ x2,
Then H (x2) - H (x1)
＝[3x2＋(m²/x2)－2m]－[3x1＋(m²/x1)－2m]
＝(3x2－3x1)＋[ m²(x1－x2) /(x1x2) ]
＝[3x1x2(x2－x1)＋m²(x1－x2)]/(x1x2)
＝[(x2－x1)(3x1x2－m²)]/(x1x2)
＝(x2－x1)[(3x1x2－m²)/(x1x2)]
∵x2－x1＞0,3x1x2－m²＞3m²－m²＞0,x1x2＞0
∴h(x2)－h(x1)＞0
That is, H (x1) < H (x2)
That is, H (x) is a monotone increasing function in [M, + ∞)