After learning the properties of the equation, Xiao Ming said to Xiao Liang that I found that 3 can be equal to 2. You see, there is an equation 3x-2 = 2x-2, where both sides of the equation are added at the same time Let's go to 2 and get 3x = 2x, and then divide both sides of the equation by X at the same time and get 3 = 2 (1). Please think about it. Is Xiao Ming right? Why? (2) can you find the solution of the equation 3x-2 equal to 2x-2?

After learning the properties of the equation, Xiao Ming said to Xiao Liang that I found that 3 can be equal to 2. You see, there is an equation 3x-2 = 2x-2, where both sides of the equation are added at the same time Let's go to 2 and get 3x = 2x, and then divide both sides of the equation by X at the same time and get 3 = 2 (1). Please think about it. Is Xiao Ming right? Why? (2) can you find the solution of the equation 3x-2 equal to 2x-2?

(!) no, because we don't know if x is 0, we can't divide X by both sides at the same time
(2) We get 3x-2x = 0 by term shifting and x = 0 by solution

If 3x-6 = 5, we get 3x = 5 +? According to the nature of the equation, what is it, on both sides of the equation?

3x = 5 + 6

The two vehicles run from 403.5 km away at the same time. After 8.5 hours, they are 38 km away. How many km does vehicle a travel 20 km per hour and vehicle B travel

(403.5-20*8.5-38)/8.5
=22.94 km / h

If △ ABC is inscribed on a circle with radius R and 2R (sin ^ 2 a-SiN ^ 2 b) = (√ 2a-b) SINB, find the maximum area of △ ABC
It has been found that C = 45 degrees, edge C = root 2 * r

2R(sin²A-sin²C)=(√2a-b)sinB
(2R)²sin²A-(2R)²sin²C=(√2a-b)*(2R)SinB
a²-c²=(√2a-b)b=√2ab-b²
a²+b²-c²=√2ab
cosC=(a²+b²-c²)/(2ab)=√2/2
C = 45 degrees
c=2RsinC=√2R
c²=2R²=a²+b²-√2ab≥(2-√2)ab…… Take the equal sign when a = b
ab≤2R²/(2-√2)=(2+√2)R²
S=(1/2)absinC=(√2/4)ab≤[(√2+1)/2]R²
That is: the maximum area of triangle ABC = [(√ 2 + 1) / 2] R & sup2; (in this case, a = b)

The length of train a and train B are 144m and 180m, and train a runs 4m more per second than train B. the two trains are facing each other, and it takes 9s from meeting to staggering. What's the speed of the two trains?

Suppose car B travels XM per second, then car a Travels (x + 4) m per second. According to the meaning of the question, we get: 9 (x + X + 4) = 144 + 180, sort out: 2x = 32, solve: x = 16, then car a travels 20m per second, car B travels 16m per second

There is a set of data 3,8,8,7,8, the average of X is exactly equal to the mode x, X should be

(3+8+8+7+8+x)/6=8
x=14

A and B walk toward each other from a and B. A walks 3 kilometers per hour, B walks 2 kilometers per hour. When they meet, they are 3 kilometers away from the midpoint. How far is the distance between a and B?

When meeting, a takes more distance than B: 3 × 2 = 6 (km), the time for meeting is: 6 ÷ (3-2) = 6 (hours), (3 + 2) × 6 = 5 × 6, = 30 (km), a: A and B are 30 km apart

-The method of finding the law of 1,2, - 4,8, - 16,32
Please make sure the method is easy to understand
And the formula of the nth number is (- 1) ^ n * 2 ^ (n-1)
How did you get it!

General term = (- 2) ^ (n-1)
That is, the latter term = the former term × (- 2)

On the map of 1:5000000, the distance between a and B is 6cm. If you draw it on the map of 1:30000000, you should draw () cm? To calculate!

6 × 5 million △ 3 million = 10 cm
Promise to draw 10 cm

Let C lie in the closed region (including the boundary) enclosed by the parabola y2 = 2x and the straight line x = 3, then the maximum value of the radius of the circle is ()
A. 32B. 4-6C. 4+6D. 6-1

When the radius of circle C is the maximum, we know from the symmetry that the center of circle C should be in the interval (0, 3) on the x-axis, and the circle C is tangent to the straight line x = 3. Let the center of circle be (a, 0) (0 ＜ a ＜ 3), then the equation of circle C is (x-a) 2 + y2 = (3-A) 2 ^, where y2 = 2x is substituted, (x-a) 2 + 2x = (3-A) 2 ^, that is, X2 + 2