Questions about learning math, questions about math?

I don't know if you have such a problem: in the past, you could get high scores without reviewing, because you usually studied thoroughly; After the third year of junior high school, I usually learn more sloppily, but I still treat the exam with my previous understanding and don't review, so it will be very difficult.

No matter what the reason is, it is recommended not to measure the current self by the previous self, but to correct the mentality, start from now on, and start from what level you are now.

Some people are swayed by the ambivalence of how they used to be, and what is wrong with them now, and they can't do it all of a sudden, and they don't want to forge ahead and struggle, but just entangle and entangle, so they are falling farther and farther away.

You can look for the reason (explain the details, ask some elders to help you unravel), but you must face the reality and work diligently while looking for the cause.

Can you be specific?

For example, how do you feel when you see the question? What about before?

What types of questions are not done well? Is it because some of the knowledge is not solid?

Don't be afraid. Thinking that you are a math strongman.

But it's also about learning with an open mind.

Ask your teacher more. I am also very good at mathematics, and I participated in the National Mathematical Olympiad Invitational Tournament held in Hong Kong, and I won it with confidence and humility.

Don't do more questions. Listen.

You have now entered the misunderstanding, I also encountered the same problem in my third year of high school, I often do a lot of questions, I hope to see more question types, but in the end I found that my mathematical thinking is dull, and finally I asked the teacher what my problem was, and the teacher told me not to do new problems, just go and see what I have done. I used to do math every day, but I suggest that you don't put too much energy into it, because if you blindly try to do well in math, there will be invisible pressure, so you focus on how to do well in the exam, and don't think about it properly.

Workaround:1. Find the wrong question book, record the wrong questions, and look through it often.

2。Learn to push the formula yourself, don't rely on memorization, rely on methods.

3。Don't take it too seriously, the more you want to get closer to him, the more it will move away from you.

4。You have the confidence to win, and strive for a perfect score, not a lot of points.

As long as you master the several important question types mentioned by the teacher, I believe you will be able to learn well. Overcome yourself. I used to be bad at math all the time.

The two arguments of the inequality are multiplied by 6 at the same time, and the direction of the inequality sign remains unchanged. Silver Elimination.

According to the two-point principle.

1.Scores. To remove the denominator, both sides of the expression should be multiplied by 6.

2.The nature of inequalities. The two sides of the inequality are multiplied at the same time with a positive number of 6, and the direction of the unequal line is unchanged.

Based on these two points, we get the following formula.

Hello, what problem do you have [Bixin] [Bixin] [Bixin] [Bixin] [Bixin] I'll see if I can solve it.

Heart] [heart] [than heart].

Questions. <>

That's the first question.

In the first question, you make a perpendicular bisector of 3 sides (a perpendicular line at the midpoint), and the intersection point of the 3 lines is the O point.

In the second question, you do the perpendicular bisector of ab, and the intersection point with the straight line l is that point.

Questions. Okay thank you.

[Heart] [Heart] [Heart] [Heart] [Heart] [Heart] that should be done if you have nothing to do

Questions. Good.

Thank you. Heart] [than heart] [heart] [heart] [heart] [heart] [celebration] questions. <>

Question 3: How to do it.

Wait a minute, [heart] [celebrate] [heart] [celebrate].

<> this is the whole process [celebrates] [hearts] [celebrates].

Questions. Ok thanks again

2+x) Clearance 2 2x-1) 3

Through observation, the denominator on the left is 2, the denominator on the right is 3, to answer the world to change back to if you want to remove the denominator on the left and right sides at the same time, you need to multiply the same number on the left and right sides at the same time, this number must also be a multiple of 2 and 3, 2 3 = 6, therefore, choose 6 is the most appropriate, the left and right sides are multiplied by 6 at the same time, because 6 0, the direction of the unequal sign is unchanged, and the original formula can be changed.

2+x)/2 ×6≥ (2x-1)/3×6(2+x)×6]/2 ≥ 2x-1)×6]/32+x)×3≥ (2x-1)×2

According to the two ends of the inequality multiplied by a positive number at the same time, the direction of the inequality does not change, and the denominator of the reed circle at both ends of the inequality is 2, 3, and the least common multiple is 6, so the two ends are multiplied by a positive number 6 when they are the same age

This is the solution of the unary one-time inequality, look at the figure below.

2+x) 2 2x-1) 3, multiplied by 6 on both sides, Xian Tan because of 6 0, therefore, the unequal sign does not serve the omen. i.e. 3(2+x) 2(2x-1).

6+3x Brother's Scatter 4x-2, after moving: 4x-3x 6+2

i.e. x 8

2+x) Fan Zheng Meng 2 (2x-1) 3

Multiply the two sides by 6

3(2+x) state bridge 2(2x-1).

6+3x≥4x-2x≤8

Choose C, which is a 1-2-3-4-5-6 square series, 1-4-9-16-25-36

As shown in the hail belt map, just do it or change the shape of the source bridge into.

There were 3 consecutive hits to the target Another 2 misses included three cases, i.e. three consecutive shots in the first, in the second, in the third, and these three cases were mutually exclusive, and the results were obtained according to the formula of probability of independent repeated trials and mutually exclusive events

Untie; Let "hit the target with the ith shot" as the event ai (i=1,2,3,4,5);

The shooter hits the target consecutively out of 5 shots, 3 times in a row, and the other 2 misses the target" for event A, then.

p(a) = (a1 a2 a3 non-a4 non-a5) + (non-a1 a2 a3 a4 non-a5) + (non-a1 non-a2 a3 a4 a5).

I hope you can understand.

Using the jack method, the probability of hitting the target three times in a row is (2 3) 3 = 8 27 as a whole, inserting two missed shots, there are three cases, so p = 3 * (1 3) * (1 3) * (8 27) = 8 81

Using the jack method, three consecutive hits to the target look at a whole, the probability is 8 27, in the other two arrangement, there are two rows, p = 3 * (1 3) * (1 3) * (8 27) = 8 81

If the edge length of the tent is r, then the pyramid height is the root number (9-r 2), so the tent volume is (3 * root number 3) * r 2 2 + (root number 3) * (root number 9-r 2) * r 2 2 = (3 * root number 3 + (root number 3) * (root number 9-r 2)) * r 2 2. When r = 2 * root number 2, the maximum value of 16 * root number 3

Let the distance from the top of 6 rhombus cone to the bottom of 6 rhombus cone be x, then the side length of the bottom of 6 rhombus cone can be calculated as 9-x 2Let the volume of the 6 cone be v, and v= - 3 2(x 3-9x) can be calculated from the above conditions. Then we derive v to get v'= - 3 2 (3x 2-9), and we can see that x is an increasing function on (- 3, 3), and v gets the maximum value when x = 3.

Question 1: 1 and 2/3 = 18/18 (30).

Question 2: A fraction smaller than 1/3 but greater than 1/4, for example, it can be 7/24 Question 3: A fraction smaller than 1/4 but greater than 1/5 can be 17/80, 18/80, 19/80

Question 4: This question can be understood from two perspectives:

1) Purely in terms of quantity, A has the most and C has the least.

2) From the point of view of production speed:

The average number of units per hour of A is 34 9=

The average number of B per hour is 31 8=

The average number of C per hour is 26 7=

So, B does the most, and C does the least.

1 and 2/3 = 18/18 (30).

A fraction smaller than 1/3 but greater than 1/4 16.

A fraction smaller than 1/4 but greater than 1/5 120.

To make the same parts, A made 34 pieces in 9 hours, B made 31 pieces in 8 hours, and C made 26 pieces in 7 hours.

A did 34 9= pieces in 1 hour, B did 31 8= pieces in 1 hour, and C did 26 7= pieces in 1 hour.

B does more, C does less.

1. 30

3. 19/80,18/80,17/804.A speed 1904 504, B speed 1953 504, C speed 1872 504, B more C less.