Phương trình vô tỷ - Bình phương 2 vế - Toán 9

The second method I use is the TIC method, which is similar to the way we analyze atoms. We can use a hidden function to analyze atoms, or we can analyze directly. First, let's look at the models we often encounter. The first model is a TIC. The first model we often encounter is a TIC. A TIC. Can you read it? What does it mean? Can you read it? What is it? 1, 2, 1, 2. Of course, if a TIC plus a total, then this place is 1 plus A and 1 plus B. This is not only applied in atomic mechanics, in programming, in programming, but also in quantum mechanics or in mathematics. This is especially important. The second model, AU plus BV minus AV plus BU. What is this? That is, normally, 4 people go together, 4 people go together, we pair together, then it will be an atom. It is A minus B. A minus B minus B. The third model we often encounter is the VS model, that is, it is X minus A plus B times X plus AB. How? Can you read it? In the past, when I was in the 8th grade, there was a table like this. Let's analyze this guy as an atom. Minus 7X plus 10 like this. X minus A plus 10 is an atom, right? Then the student will take 10 and divide it by 2 times 5. And 7 divided by 2 plus 5. Once that is done, this is a total. This is a total like A plus B. And 10 is a TIC like A plus B. Then immediately it will appear. The atom is X minus 2 times X minus 5. The student will know right away. This is X minus 2 times X minus 5. So it's the same. Above will be X minus A times X minus B. After that, just meet a classmate in class 2. Then people will observe that the free number that people analyze becomes an atom. And the number of X, people check to see if it is a total of 2 atoms to analyze or not. If so, immediately people will have atoms like this. So now back to my number 5. When people solve the equation, they look at the number 1. And this is the TIC. It is the TIC of these two atoms. These two atoms multiply each other by this. So it is exactly this model. And because of that, the equation number 5 will always be equal to number 5. Here you have to say the formula, not the equation 7 anymore. The equation will always be equal. According to my theory, this is a TIC with a total. Then it will become an atom. 1 minus this atom. 1 minus X minus 2. Multiplied by 1 minus X plus 2 like this. And of course, what is this? It is our model 1. I will simplify a bit. Here, in this number 6, it has a note like this. First, it looks a bit big. Then I ... This number 5 is a form called 1 plus a TIC. A total TIC. And number 6. In number 6, we have a note like this. When meeting the 2-channel sign, then try to multiply by the sum. That is, it is a sign. When meeting the 2-channel sign, then try to multiply by the sum. That is, it is a sign. Then I will multiply by the total. Have you ever heard of multiplying by the sum? Not yet. Here, multiply by the sum like this. I will have ... If I meet a friend who is A minus B like this. Then I will write it as ... I multiply by the sum. This is a sign. Then I multiply by the sum. A plus B. I multiply by the sum. Read it. So what is it equal to on the board now? According to the formula, what is it equal to? A minus B. The formula is still the same, right? So later, when I meet this friend, I will multiply by the sum and it will be this value. But in practice, I should not multiply. And I will replace this by this. And this replacement is called multiplying by the sum. Of course, these equations need to have conditions. Sometimes I don't put conditions because I'm here to discuss my theory only. The equation is another thing. The condition is ... The condition of this equation is that x will be greater than 1. But when I calculate, the first condition is the condition first. Condition x is greater than 1. Because sometimes, the condition of the equation calculates many things. When x is greater than 1, this column is greater than 0. And these columns are all equal. And this big column. Usually, this big column in the equation has 3 columns. Then the big column will be the sum of 2 small columns. Usually it is like that. Usually there will be such a situation. So then ... Now I will replace ... In this case, I can understand that ... This is another model. That is, what is A minus B equal to? With the previous model, what is A minus B equal to? Did I get it wrong? Can you guess? It will be equal to ... K minus B. K minus B is equal to A plus B. That's right. It will be A plus B. And this action is called the union action. But I remember. I always remember. I always use it to practice quickly. But in this case, I can say something like this. We have this column plus this column. The column of x plus 2 plus the column of x minus 1 is greater than 0. When it is greater than 0, then I am multiplied by 2 and it is equal. Because A is equal to B. It is equal to AC equals BC. But with the condition that C must be square. C must be square or C must be negative. Or C must be different from 0. But how do you know if it is different? When it is always square or always negative, will it be different? So in this model, if it is always square, I am allowed to multiply. Or if it is always negative, I can also multiply. But I have to point out that it is always square. I point out that it is always square. So at this point, the equation is equal. The equation means my equation 6. The equation is equal. I will multiply this. I multiply by 2. I multiply by here. This action is called the multiplication action. So when I meet a two-column symbol, I try to multiply. In some cases, I meet a two-column symbol. I also multiply, but I will multiply negative. So when I have this symbol, I multiply this column. What will it be? It will be this. This symbol multiplies this column. It will be like this. I just multiply the column by 2. It will be equal. What will it be? There is a funny thing that happens. What will it be? This becomes what? It is number 3. This number 3 and this number 3 are missing. So this must be 1 plus the sum. 1 plus the sum is my familiar form. It will be 1 minus d tan of x plus 2 times 1 minus d tan of x minus 1. Is it right? In this problem, I can't analyze this formula. But through the two-column symbol, I multiply it, it will give me the formula. This problem is very simple. Have you seen it yet? Do you see it simple? What is this equation? 6 minus d tan 6 minus 3b What is this equation? a minus 3 times b minus 2 Just put them together. Or not. Just put them together. Or this and this. There are 4 of them. Just put them together. It is the form of au plus bv equals av plus bu. It is that formula. Let's look at this equation. 3x minus 1 plus 3 tan of x minus 1 plus 3 tan of x This equation is similar to the equation of nature. It can be said that it is similar to the equation of nature in the second round. Of course, it is in the formula to give the formula. But before we can give the formula, we need to make it beautiful. What is it? We can use to make it beautiful by using two constants plus two tan as constants. But before this equation, the condition is not simple. The condition is not simple. Namely, in terms of conditions, tan of x must be greater than 0. x plus 3 times x minus 1 must be greater than 0. In this case, if you are not careful, you will get it wrong. In this case, students are good. The first time is also easy to get it wrong. Here, it is wrong like this. You see, x plus 3 is greater than 0. Equally, x plus 3 is greater than 0. x minus 1 must be greater than 0. Is it reasonable? I repeat, x plus 3 is greater than 0. For x to be greater than 0, x minus 1 must be greater than 0. Therefore, x plus 3 must be greater than 0. And x minus 1 must be greater than 0. When we solve this equation, x minus 1 must be greater than 0. Therefore, x plus 3 must be greater than 0. But this condition is wrong. And where is it wrong? Where is it wrong? Is this situation wrong? Where is it wrong? There is a saying like this. Self-righteousness is like a pen. It means thinking right or wrong. There are many situations where I think it's right but it's not. It's vague. The reason is like this. If a plus b is greater than 0 and a is greater than 0, we cannot conclude that b is greater than 0. Can you explain why? Do you understand why if a plus b is greater than 0 and a is greater than 0, we cannot conclude that b is greater than 0? Why is that? Because when a is greater than 0, then b will be arbitrary, right? If a is greater than 0, then b can be negative, right? b can be negative. b can be positive. In general, it's arbitrary, right? When a is greater than 0, then it doesn't matter which one is greater than 0. They are all equal, right? Therefore, when we encounter this situation, we will set two situations. We will set two situations. One is a is greater than 0. At that time, b is arbitrary. The other is a is positive. If a is greater than 0, then b is positive. At that time, what will happen to b? If a is positive, then what will happen to b? b is greater than 0. Because we can divide a by 0. Actually, this situation has to be three situations. But in my situation, I have a is greater than 0. That's right. So if it's true, it has to be like this. It has to be equivalent to the case x is greater than 0. If x is greater than 0, then it's OK. That is, x is equal to negative 3. And the second case is x is greater than negative 3. That is, x is equal to negative 1. At that time, x is equal to 1. If we perform an addition, it will be like this. Negative 3 is here. Negative 1 is here. But for the negative 3, it will be like this. And from 1 to 0, it will be like this. It's like that. So that's its condition. So if you are a good student, the first time you meet this case, you will get the wrong condition. So now, this form and this formula are used for natural multiplication. So now, with such conditions, I will set x is equal to negative 3. x is equal to negative 3, is it satisfied? It's not satisfied, right? Then eliminate it. With such conditions, x is equal to negative 3, then eliminate it. So I set x is equal to 1. I set x is equal to 3 to a. What is the condition for a? When I perform an addition, remember to set the condition for it. a is equal to 0. a is equal to 0 is not satisfied. x is equal to 2. x is equal to 2, then b is equal to 0. With such conditions, I will solve it later. a is equal to 1, a is equal to 3.2, I eliminate it, I don't have to do anything else. So now it has a requirement, that is, we have performed the addition, but the 3x minus 1, we need to perform the 3x minus 1 according to a, b. We have to perform it according to a, b. At that time, there will be a process of a and b. At that time, we can analyze it into a function, right? But how do we perform it? Let's do it like this. We can observe x is equal to 3, a is equal to x plus 3, b is equal to x plus 1, that is, the numbers are all 1. At that time, there must be one person, there are two times one person. So I will perform it like this. a plus 2b, let's try it like this. a plus 2b, what will it be? Read it. 2b plus 2b, what will it be? 3 plus x 3x into 3x plus 1 plus 1 So we have a process of transformation. At that time, the process will be transformed. Don't think it's an equation, but a process of transformation. I have to remind you about the equation. There is an idea that the equation is defined as two equal processes if their experiences are the same. So if x is a, b, or any other function, I won't say it's an equation, I will say it's a process of transformation. A process of transformation. Or in general, it's a process of transformation. So it will become a plus 2b minus 2, right? Here, it must be minus 1. Plus 3ab minus a equals 0 equals a. So it's equal. Now I have to analyze this. Do you remember this formula? a square plus 3ab plus 2b square minus a minus 2 equals 0 This is a formula. The formula we use is the formula that we use This is a formula. Do you know how to do it? Do you know? The requirement of a math teacher is that they know how to do it. Because it has a formula and a method. The formula is ax2 axx plus b di2 plus di plus c di2 plus ax plus bi plus c I will be careful. I have some thoughts like this. The first step is to analyze the second group of elements. The analysis of the second group of elements is u times v. It is u times v. In the second step, I will represent the second group by u times v. That's it. Do you remember this formula? I think you remember. When you meet this formula, the first thing you observe is that this is the second step. This is the first step. This is the second step. In the first step, I will analyze the elements. There is good news. If you can analyze the elements in the second step, you will have a chance to analyze. If you can't analyze the elements in the second step, you will never be able to analyze the elements. This is very important. Because if you know that you can analyze the elements, you will have a chance to continue. If you see this example, ax2 axb2 and you know that you can't analyze the elements, you will fail to analyze the elements. That's it. You can evaluate whether it is possible or not. I see that it is possible. Specifically, this is axb times ax2b And this friend a I will represent these two. In fact, it will be 2 of I multiply 2 by 2. It will be like this. 2 of axb minus ax2b This friend a I will represent this minus this. Let's say this is u and this is v. In this step, I will represent u and v like this. 2u minus v like this. And here it is not done yet. So here it becomes u times v plus 2u minus v minus 2 And so what? u minus u minus 1 u minus 1 or u plus 1? u minus 1 plus 2 u minus 1 times v plus 2? Ah, it looks like I'm wrong. It's wrong here. It must be the opposite. So u plus 1 times v minus 2 It's wrong. Because my u is axb, so it will be negative. So u plus 1 will be negative. And because of that, it will be negative, so it's just v times 2. In short, the next step is easy. So look back a little bit. There is a method to bring it back to negative. But there are tables that bring it back to negative in a simple way, such as 7. If you bring it back to negative, it won't come out. But this table 8 also brings it back to negative, but not right away. We have to pass the hidden function. For example, because it has two bars, so I use two hidden functions. But I will have a problem after the hidden function is done. After the hidden function ab is done, it must be able to perform this outside. Perform this triplet. According to that a and b. But it has many performances. So we have to try some performances. For example, in this case, the number here is 3. But the number of x here is 1. There must be one that becomes 2. So in this case, I try that 2b squared is successful. But if I try 2a squared plus b squared, it is not successful. If it's not successful, I have to try again. And have to be persistent and confident. Persist on the path I have chosen. This section is a section for the second round. These are the difficult sentences in the exam. Actually, there are not many difficult sentences in this course. It's just like this. This is the method for the test. There are about 20 methods in this course. I have learned only 2 methods tonight.