Vector and triangle In triangle ABC, given that vector AB and AC satisfy {(AB / AB) + (AC / AC)} * BC = 0, what triangle is triangle ABC? (AB and AC are vectors) AB / (AB) = unit vector in AB direction, AC / (AC) = unit vector in AC direction, and + is unit vector in bisector direction of angle AB and AC Why do the two add up to the unit vector in the direction of bisector of angle AB and AC? What is the meaning of "unit vector in AB direction, unit vector in AC direction"?

Isosceles triangle. The AB unit vector and AC unit vector are set as am, and the baseline is the bisector of angle a, and am is perpendicular to BC, so the triangle is isosceles triangle
AB unit vector and AC unit vector are vectors of unit module length in their direction. Since module lengths are equal, they are added according to the parallelogram rule. Am is a diamond diagonal, which is naturally the bisector of angle A

The content of triangle rule: connect two vectors head and tail, and form the triangle of () with their resultant vector, so as to find the resultant vector
Two vectors are connected end to end, and their resultant vectors form the triangle of (), thus the resultant vector is obtained

Two vectors are connected end to end and form a triangle with their resultant vectors, thus the resultant vectors are obtained

Calculation (M + 1) (M & # 178; + m-1)

Solution (M + 1) (m ^ 2 + m-1)
=m^3+m^2-m+m^2+m-1
=m^3+2m^2-1

11/20x+32=3/4x+24

x=40..

What is the minimum value of quadratic function y = (x ^ 2-1) + 2?

The minimum value of x ^ 2 is 0
So y min = 0-1 + 2 = 1

Let the perimeter of a square be x and the radius of its circumcircle be y, then the functional relation between Y and X is thanks

Let the perimeter of a square be x and the radius of its circumscribed circle be y, then the functional relation between Y and X is
The diagonal length of a square = the diameter of its circumcircle
2 * (perimeter / 4) ^ 2 = (diagonal length) ^ 2
Diagonal length = √ 2 * (perimeter / 4) = (x / 4) * √ 2
Diameter = 2 radius = 2Y
(x/4)*√2=2y
y=(√2/8)x

A + B + C = 0 ABC is not equal to 0 find the value of a (1 / B + 1 / C) + B (1 / C + 1 / a) + C (1 / A + 1 / b)

Solution: a (1 / B + 1 / C) + B (1 / C + 1 / a) + C (1 / A + 1 / b)
=a/b+a/c+b/c+b/a+c/a+c/b
=(a+c)/b+(b+c)/a+(a+b)/c
=-c/c-a/a-b/b=-3

How to solve the equation 2x-1 / 2 = 4x Square-1 / 4?

2x-1 / 2 = 4x ^ 2-1 / 4 4x ^ 2-2x + 1 / 4 = 0 (the left is the complete square) (2x-1 / 2) ^ 2 = 0 2x-1 / 2 = 0 2x = 1 / 2 x = 1 / 4

How to solve the equation (2x 3) / 3-1 = - ((5x-6) / 4)

Take 12 on both sides
4(2x+3)-12=-3(5x-6)
8x+12-12=-15x+18
8x+15x=18
23x=18
x=18/23

Given the circle C: x + y-4x-14y + 45 = 0, P (x, y) is any point on the circle, find the range of (Y-3) / (x + 2)
The square of x the square of Y

First, the circle is transformed into a standard circle, that is, (X-2) &# 178; + (Y-7) &# 178; = 0
Let (Y-3) / (x + 2) = t, then y = t (x + 2) + 3
If y = t (x + 2) + 3 and circle (X-2) &# 178; + (Y-7) &# 178; = 0 are solved simultaneously, the problem of t can be solved, that is, the range of (Y-3) / (x + 2), so that the line and ellipse have focus

The content of triangle rule: connect two vectors head and tail, and form the triangle of () with their resultant vector, so as to find the resultant vector Two vectors are connected end to end, and their resultant vectors form the triangle of (), thus the resultant vector is obtained