Problem solving thinking in high school math commutation

When we use the commutation method, we should follow the principle of being conducive to operation and standardization, and we should pay attention to the selection of the range of new variables after the commutation, and we must make the range of new variables correspond to the value range of the original variables, and cannot be narrowed or expanded. As in the above examples, t>0 and [0,.

You can observe the equation first, you can find that there are always the same formulas in the equation to be commuted, and then replace them with a letter, calculate the answer, and then if there is this letter in the answer, you can bring the formula in, and the calculation will come out.

If you encounter the form of x y 2s, set x

s＋t，y＝

s t and so on.

Triangular exchange of dollars. When it is applied to remove the root number, or when it is transformed into a trigonometric form, it mainly uses a certain connection with the trigonometric knowledge in the known algebraic formula for commutation. For example, when finding the range of the function y 1-x 2, if x [-1,1], let x sin

sin [1,1, the problem becomes the familiar domain of trigonometric functions. The main reason why I thought of such a setting should be to discover the connection of the value range, and there is a need to remove the root number. For example, the variables x and y fit the condition xyr

r>0), it can be used as a trigonometric substitution x rcos and y rsin to turn it into a trigonometric problem.

Is there any question can hi me.

1. The question of changing the yuan is not changed for the yuan, and it is done!

2. Swap: Look for similar expressions to replace them with a letter.

Just look for expressions like that!

Pay attention to the change: define the domain.

For example, if we set the root number under (-x*x+6x)=t, the definition field of t is [0,3], because the root number is greater than zero, and the maximum value of the quadratic function in the root number is 9

Common commutation: The most common quadratic substitution you use right now: replacing a function with a quadratic function.

1》f(x)=4^x2^x

In high school mathematics, the commutation method is a commonly used method to solve analytical formulas. It simplifies the solution of the problem by introducing new variables or by deforming the original equations. When using the commutation method to solve the analytic formula, the steps of common slag are as follows:

1.Determine the appropriate variable substitution such as Lingqi: By observing the characteristics and properties of the original equation, select a suitable variable substitution.

The purpose of variable substitution is to transform the original equation into a simpler form. 2.Perform variable substitution:

After the selected variable is replaced, the unknown quantity in the original equation is represented by the new variable and the variable is replaced. This gives you a new equation. 3.

Solving new equations: According to the characteristics and properties of the new equations, the new equations are solved by using known mathematical methods, such as equation operation, equation solving, etc. If the analytic formula of the new equation can be obtained, the solution of the commutation method is completed.

4.Solving the original equation: By reverse substitution, the new variables in the new equation are represented by the original unknown quantities, so as to obtain the analytic formula of the original equation.

It should be noted that the commutation method is widely used in high school mathematics, and the common commutation methods include direct substitution, homogeneous substitution, double angle substitution, parameter substitution, etc. Which commutation method to use depends on the specific solution problem. For example:

Suppose the equation is required to be solved x 2-2x + 1 = 0. It is observed that this equation can be solved in the form of a squared completion. 1.

Variable substitution: let y=x-1, convert the original equation to y 2= solve the new equation: from the new equation y 2=0, we can see that y= solve the original equation:

According to the relation of variable substitution, y=x-1 is brought back to the original equation, and x-1=0 is obtained, and x=1 is solved. Therefore, the analytic formula of the original equation is x=1. It should be noted that the commutation method is only one of the commonly used methods in solving mathematical problems, and the process of solving the analytical formula also needs to be analyzed and judged in combination with specific problems, and the appropriate methods and approaches should be selected.

Let's take an example.

f(1 x)=x, we know that x≠0, that is, the value range and the definition domain of the function are both a set of x≠0. If y=1 x, then there is x=1 y, and if you bring in the original function, it is f(y)=1 y, and because the selection of the independent variable in the expression of the function is careful, it can be written as f(x)=1 xTo sum up, the commutation method should first consider the value range of the function definition domain, and the family base should not change because of the change of the definition domain and value range after the conversion, and then consider the overall substitution of the independent variables, and finally adjust the form.

Please point out the shortcomings.

This question is indeed easier to understand.

First of all, we generally evaluate the range and define the domain, which is transformed from a function of a known kind.

For example, in the function y=x 2, the definition range is negative infinity to positive infinity, and the value range is y>=0. This is equivalent to what we know.

In the title, y=1 (x 2+3) is not the type of function we know. Then we'll use the commutation to transform it into a familiar type.

Set: t=x 2+3.........1）

then the original function becomes.

y=1/t………2）

These two functions are obviously well-known functions. In this way, the original function becomes the two function types that we are familiar with, and we can use the knowledge of the relevant definition domain range of these two function types to find the value range of the original function.

There is a problem with understanding the teacher's words, the teacher is saying that after the change of elements, t will also have a range, and in this range, the range of y is the range of the original function......Got it? This is just a ...... that makes use of the equivalence principle of algebraFor a simple example: (a+b+c) -2(a+b+c)+1=0, find a+b+c, we will make t=a+b+c, solve the quadratic equation about t, and find t is a+b+c......

It's better to practice on your own.

The teacher's meaning can be understood as let x+1=t, then x+3=t+1, so f(t)=t+2, and since t and x are both variables, x(x)=x+2 can be used instead

In fact, the term is the same as the original idea, you just need to understand the x here, x is just a variable, just an unknown, he can replace any number, here x and x+3 or x+1 are just variables, where x is not the same, f(x+1)=x+3 = (x+1) +2 so f(x)=x+2 where x+1 can be substituted, this is just a transformation of the idea The equation can be replaced by any unknown, such as y z t a .

It's just that we're used to x.

If you go long, you will experience it.

Make-up is to turn the original formula into the form you need.

For example, this question: because f(x) is x+1 in parentheses and x+3 on the outside, so if you want to be consistent inside and out, replace x+3 with (x+1)+2, in fact, the expression (x+1) +2 on the outside is equivalent to x+3, and your idea actually changes the function.

To make up an item is to make an item according to the need and subtract it later;

For example, x +2x=x +2x+1-1=(x+1) -1 Think about it, do you often use this method when solving a translation problem?

The change of yuan, the real question for you, is to replace f(x+1) with the form f(x).

You can't do that, the commutation method is to replace one thing with several other things, which plays the role of simplification, for example, y=a+z+s+x b=z+s+x to get y=a+b

This does not require a change of yuan, which is to replace a complex formula with a simple formula.

f（x+1)=x�0�5+2x+31.Let the number in f() parentheses be another value and set x+1=t2variant, x=t-13

Bring in the algebraic formula of x into the analytic formula f(t)=(t-1) 0 5=2(t-1)=3 f(t)=t 0 5+24Swap the element and replace the t in f() parentheses with x f(t)=t 0 5+2, that is, f(x)=x 0 5+2 Note When replacing the analytical formula of x with t, pay attention to the value range. This question is not required, but some questions do.

For example, when replacing 1 x with t, it should be noted that x is not equal to 0, so 1 x is not equal to 0, that is, t is not equal to 0Therefore, it should be written like this, so that when you write 1 x=t(t≠0) and finally write the analytic expression of x, you should also write the value of x. For example, in the above example, the analytical formula of f(x) is f(x)=x 0 5+5, and the value range of (x≠0) should be written at the end of the value of x, which should be the same as the value of the set t value.

Let t=x+1x=t-1 bring in the function to get f(t)=(t-1) 0 5+2(t-1)+3 =t 0 5+2, so f(x)=x 0 5+2