First, we decompose the factor in the evaluation: (B-A) square + B (a-b), where a = 2, B = - half

First, we decompose the factor in the evaluation: (B-A) square + B (a-b), where a = 2, B = - half

（b-a）²+b(a-b)
=（b-a）²-b(b-a)
=(b-a)(b-a-b)
=-a(b-a)
=-2*(-1/2-2)
=5

3 (x-3) = (x-3) &# 178; 4x & # 178; - (x-1) &# 178; = 0 is factorized
Starting from 4, it's another way

1)3(x-3)=(x-3)²
(x-3)²-3(x-3)=0
(x-3)(x-3-3)=0
x1=3 x2=6
2)4x²-（x-1）²=0
(2x+x-1)[2x-(x-1)]=0
(3x-1)(x+1)=0
x1=1/3 x2=-1

(X-Y) a - (X-Y) is factorized by the method of group decomposition

(x-y)a-(x-y)
=(x-y)(a-1)

Factorization of xy-x-y + 1 by group decomposition

xy－x－y＋1
=x(y-1)-(y-1)
=(y-1)(x-1)

Fill in the appropriate prime numbers within 50 in the brackets below
（ ）+（ ）+ ( )=51 ( ) +（ )+( )=61
（ ）+（ ）+ ( )=51 ( ) +（ )+( )=61
（ ）+（ ）+ ( )=51 ( ) +（ )+( )=61
（ ）+（ ）+ ( )=51 ( ) +（ )+( )=61
（ ）+（ ）+ ( )=51 ( ) +（ )+( )=61
（ ）+（ ）+ ( )=51 ( ) +（ )+( )=61

（5）+（ 17）+ (29 )=51 ( 3 ) +（ 11 )+( 47 )=61
（ 3 ）+（ 5 ）+ ( 43 )=51 ( 7) +（ 11 )+( 43 )=61
（ 3 ）+（ 7 ）+ ( 41 )=51 ( 3 ) +（ 17 )+( 41 )=61
（ 3 ）+（ 17 ）+ (31 )=51 (7 ) +（ 13 )+(41 )=61
（ 3 ）+（ 19 ）+ ( 29 )=51 ( 5 ) +（ 13 )+( 43 )=61
（ 3 ）+（ 11 ）+ ( 37 )=51 ( 7 ) +（ 17 )+( 37 )=61

Let the probability density of random variable X be f (x) = CX ^ 2, x > 0; 0, others
Try to find: 1 constant C; 2. E (x), D (x); 3. P {x-e (x)|

(1).1=∫[-∞,+∞]f(x)dx=∫[-2,+2]cx^2dx=16c/3,c=3/16.(2).EX=∫[-2,+2]x*(3/16)x^2dx=(3/16)∫[-2,+2]x^3dx=0.E(X^2)=∫[-2,+2]x^2*(3/16)x^2dx=(3/16)∫[-2,+2]x^4dx=(3/16)*(64/5)=12/5.DX=E(X^2)-(EX)^2=12/5....

Given the vector M = (2cos (ω / 2), 1), n = [cos (ω / 2) x, cos {(ω x) + (π / 3)}] (where ω > 0), the function f (x) = m · n, and its minimum positive period is π
(1) Finding the value of ω
(2) In the acute angle △ ABC, a, B and C are the opposite sides of angles a, B and C respectively. If f (a) = - (1 / 2), C = 3, the area of △ ABC is 6 √ 3, find the area of circumcircle of △ ABC

Vector M = (2cos (Wx / 2), 1)
(1)f(x)=2[cos(wx/2)]^2+cos(wx+π/3)
=1+coswx+(1/2)coswx-(√3/2)sinwx
=1+(3/2)coswx-(√3/2)sinwx
=1+√3cos(wx+π/6),
Its minimum positive period is 2 π / w = π, w = 2
(2)f(A)=1+√3cos(2A+π/6)=-1/2,cos(2A+π/6)=-√3/2,
2A+π/6=5π/6,A=π/3,C=3,
The area of ℅△ ABC = (1 / 2) * 3B * (√ 3 / 2) = 6 √ 3, B = 8,
According to the cosine theorem, a ^ 2 = 64 + 9-24 = 49, a = 7,
The radius of circumscribed circle r = A / (2sina) = 7 / √ 3,
The area of circumscribed circle of ℅△ ABC = π R ^ 2 = 49 π / 3

Divide 30 into prime factors______ ．

The quality factor of 30 is 30 = 2 × 3 × 5, so the answer is 30 = 2 × 3 × 5

If for any non-zero rational numbers a and B, the operation is defined as a ⁃ B = AB + A, then (- 5) ⁃ (- 4) ⁃ (- 3)
What's the result of that

（-5）※（-4）※（-3)
=[﹙－5﹚×﹙－4﹚＋﹙－5﹚]※（-3)
=15※（-3)
=15×﹙－3﹚＋15
=－30

Given the line segments a and B (a is greater than B), draw a line segment AB so that it is equal to 1 / 2 (3a-b), and explain it

Draw a straight line
Then take AB = BC = CD = a
Then ad = 3A
Then reverse de = B
That is, e is between C and D
Then AE = 3a-b
Then make AE vertical bisector, and the intersection of AE and AE is f
Then AF = 1 / 2 (3a-b)