If vector AB = vector E1-E2, vector BC = vector 2e1-8e2, vector CD = vector 3E1 + 3e2, prove that a, B and C are collinear

If vector AB = vector E1-E2, vector BC = vector 2e1-8e2, vector CD = vector 3E1 + 3e2, prove that a, B and C are collinear

Vector BD = vector BC + vector CD = vector 5e1-5e2
=5 (vector E1-E2) = 5, vector AB,
So the vector BD and the vector AB are collinear,
A, B and D are collinear

If (vector a + vector b) ⊥ vector a, find the angle between vector a and vector B

X = the angle between a and B
(a+b).a =0
|a|^2 + |a||b|cosx = 0
1+2cosx=0
cosx= -1/2
x =120°

1+(-3)+(-5)+7+9+（-11）+（-13）+15+···+2009+（-2011）+（-2013）+2015 =

1+(-3)+(-5)+7+9+（-11）+（-13）+15+···+2009+（-2011）+（-2013）+2015
=【1+(-3)+(-5)+7】+【9+（-11）+（-13）+15】+···+【2009+（-2011）+（-2013）+2015】
=0+0+…… +0
=0

Given the function f (x) = - 3x * - 3x + 2x-m = 1, when the value of real number m is, the function has two zeros, one zeros and no zeros
Question 2: if a function has exactly one zero at the origin, find the value of the real number M

Two zeros, one zero, no zero
The difference is that you need to find the right part of the equation = 0. The equation has two solutions, one solution and zero solution
So as long as B ^ 2-4ac > 0 = 0