Let a = (x1, Y1), B = (X2, Y2) and a, B not equal to 0 1. The method of finding the collinear and co directional coordinates of a and B 2. The method of finding the collinear and reverse coordinates of a and B

Let a = (x1, Y1), B = (X2, Y2) and a, B not equal to 0 1. The method of finding the collinear and co directional coordinates of a and B 2. The method of finding the collinear and reverse coordinates of a and B

1.x1*y2=x2*y1
2.x1*y2=-x2*y1

How to use two methods to decompose the polynomial XY (XY + 4) - 2 (x + y) (x's degree 2 - XY + Y's degree 2)
As above, urgent

The first type: XY (XY + 4) - 2 (x + y) (X & sup2; - XY + Y & sup2;)
=x²y²＋4xy-2(x³-x²y+xy²+x²y-xy²+y³)
=x²y²＋4xy-2x³-2y³
=(x²-2y)(-2x+y²)
Second:
xy(xy+4)-2(x+y)(x²-xy+y²)
=x²y²＋4xy-2(x+y）[（x＋y）²-3xy]
=x²y²＋4xy-2(x+y)³+6xy(x＋y)
=x²y²＋4xy-2(x³+3x²y+3xy²+y³)+6x²y+6xy²
=x²y²＋4xy-2x³-2y³
=(x²-2y)(-2x+y²)

It's going to take 300 channels, add, subtract, multiply, divide, not mix, one step

Only 100 lanes 1.45 + 15 × 6 = 135
2.250÷5×8=400
3.6×5÷2×4=60
4.30×3+8=98
5.400÷4+20×5= 200
6.10+12÷3+20=34
7.(80÷20+80)÷4=21
8.70+(100-10×5)=120
9.360÷40= 9
10.40×20= 800
11.80-25= 55
12.70+45=115
13.90×2= 180
14.16×6= 96
15.300×6= 1800
16.540÷9=60
17.30×20= 600
18.400÷4= 100
19.350-80= 270
20.160+70=230
21.18-64÷8= 10
22.42÷6+20=27
23.40-5×7= 5
24.80+60÷3=100
25.41+18÷2= 50
26.75-11×5= 20
27.42+7-29= 20
28.5600÷80=70
29.25×16= 400
30.120×25= 3000
31.36×11= 396
32.1025÷25=41
33.336+70= 406
34.25×9×4= 900
35.200-33×3= 101
36.3020-1010=2010
37.12×50= 600
38.25×8= 200
39.23×11= 253
40.125÷25=5
41.4200-2200=2000
42.220+80= 300
43.20×8×5= 800
44.600-3×200=0
45.20+20÷2= 30
46.35-25÷5= 30
47.36+8-40= 4
48.2800÷40=70
49.98÷14 = 7
50.96÷24 = 4
51.56÷14 =4
52.65÷13 = 5
53.75÷15 = 5
54.120÷24 =5
55.200÷25 = 8
56.800÷16 = 50
57.840÷21 =40
58.560÷14 = 40
59.390÷13 = 30
60.600÷15 =40
61.72÷24 = 3
62.85÷17 = 5
63.90÷15 =6
64.96÷16 = 6
65.78÷26 = 4
66.51÷17 =3
67.80÷40 = 2
68.100÷20 = 5
69.100÷4 =25
70.240÷40 = 6
71.920÷4 = 230
72.300÷60=5
73.64÷2 = 32
74.64÷4 = 16
75.50÷5 =10
76.60÷8 = 7.5
77.96÷4 = 24
78.90÷6 =15
79.400+80 = 480
80.400-80 = 320
81.40×80 =3200
82.400÷80 = 5
83.48÷16 = 3
84.96÷24 =4
85.160×5= 800
86.4×250= 1000
87.0×518= 0
88.10×76= 760
89.36×10=360
90.15×6= 90
91.24×3= 72
92.5×18= 90
93.26×4= 74
94.7×15=105
95.32×30= 960
96.40×15= 600
97.60×12= 720
98.23×30= 690
99.30×50=1500
100.5×700=3500

Given that the sum of solutions X and y of the equations X-Y = 2A, x + 3Y = 1-5a is a positive number, the value range of a is obtained
X-Y = 2A (1) x + 3Y = 1-5a (2) how to get 2x + 2Y = 1-3a, how to add (1) and (2)

(1)+(2)
Add left and left, add right and right
x-y+x+3y=2a+1-5a
2x+2y=1-3a

Use the nine numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 to form three three digit numbers (each number is used only once). The sum of the three three digits is the largest______ ．

9 × 100 + 8 × 100 + 7 × 100 + 6 × 10 + 5 × 10 + 4 × 10 + 1 + 2 + 3 = 900 + 800 + 700 + 60 + 50 + 40 + 3 + 2 + 1 = 2556 A: the maximum sum of these three digits is 2556

Given the hyperbola C: x ^ 2-y ^ 2 = 1, the slope of the line L is equal to?

a²=1
b²=1
So B / a = 1
So the slope of the asymptote is ± 1
There are two intersections with the right branch
You can see it by drawing
Parallel to the asymptote, there is an intersection
So if there are two intersections, then K1

Three students have done a total of 20 math problems. If B does one more problem, it will be twice as many as a, and a does five less than C. how many math problems have each of the three students done?

A made (20 + 1-5) / (1 + 2 + 1) = 4 (DAO)
B made 4 * 2-1 = 7
C made 4 + 5 = 9 (DAO)

If the sum of the coefficients of the quadratic equation AX2 (square) + BX + C satisfies a + B + C = 0, find the root of the quadratic equation

a+b+c=0
b=-a-c
So ax & sup2; + (- A-C) x + C = 0
ax²-ax-cx+c=0
ax(x-1)-c(x-1)=0
(ax-c)(x-1)=0
x=c/a,x=1
I wish you a happy study

How to add and subtract rational numbers 4.75-5 and 5 / 6-13 and 3 / 4 + 3 / 6

4.75-5 5 / 6-13 3 / 4 + 3 5 / 6
=(4.75-13.75) + (3 and 5 / 6-5 and 5 / 6)
=-9-2
=-11
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The quadratic equation of one variable x square + (4-2m) x + 2m square - 4M-2 = 0 has two real roots to find the maximum and minimum of the product
The maximum and minimum of the product of two

From X & sup2; - (4-2m) x + 2m & sup2; - 4M-2 = 0, we can get the discriminant: B & sup2; - 4ac = (4-2m) & sup2; - 4 * (2m & sup2; - 4M-2) ≥ 0, we can get: 24-4m & sup2; ≥ 0m & sup2; ≤ 6, that is: √ 6 ≤ m ≤ √ 6, the product of two is equal to: C / a = 2m & sup2; - 4M-2