When solving a mathematical problem, we regard a formula as a whole and replace it with a variable, thus simplifying the problem. This is called substitution. The essence of substitution is transformation, the key is to construct elements and set elements, and the theoretical basis is equivalent substitution. The purpose is to change the research object, move the problem to the knowledge background of the new object, standardize non-standard problems, simplify complex problems and become easy to deal with.

Substitution method is also called auxiliary element method and variable substitution method. By introducing new variables, scattered conditions can be linked, implicit conditions can be revealed, or conditions can be linked with conclusions. Or turn it into a familiar form to simplify complicated calculation and derivation.

It can transform high order into low order, fraction into algebraic expression, irrational expression into rational expression, transcendental expression into algebraic expression, and has a wide range of applications in the study of equations, inequalities, functions, sequences, triangles and other issues.

The substitution methods include: local substitution, triangle substitution, mean substitution and so on. Local substitution, also known as global substitution, means that an algebraic expression appears many times in the known or unknown, and it is replaced by a letter to simplify the problem. Of course, sometimes it is discovered through deformation. For example, to solve the inequality: 4+2-2 ≥ 0, first transform it into 2 = t(t >；; 0), and familiar with solving quadratic inequalities and exponential equations.

Triangular substitution is used to remove the root sign, or when it is easy to find the triangular form, it is mainly to use a certain point in the known algebra for substitution. If the range of function y =+ is found, it is easy to find x∈[0, 1]. Let x = sin α, α ∈ [0,0], and the problem becomes the range of trigonometric functions that everyone is familiar with. Why did you think of this setting? Mainly to find the relationship between the range of values and the need to remove the root sign. For example, the variables x and y apply to the condition x+y = r (r >; 0), x = rcos θ and y = rsin θ can be transformed into trigonometric problems by trigonometric substitution.

It means substitution, such as x+y = s form, let x =+t, y =-t and so on.

When using substitution method, we should follow the principle of facilitating operation and standardization. Pay attention to the selection of the new variable range after substitution, and make sure that the new variable range corresponds to the value range of the original variable, which cannot be reduced or expanded. T>0 and α∈[0,].

Ⅰ. Reproducible problem group:

The maximum value of 1.y = sinx cosx+sinx+cosx is _ _ _ _ _ _ _.

2. let f(x+1) = log (4-x) (a >1), then the range of f (x) is _ _ _ _ _ _ _ _.

3. Given the series {a}, a =- 1 and a a = a-a, the general term of the series is a = _ _ _ _ _ _ _ _ _

4. Let the real numbers x and y satisfy x+2xy- 1 = 0, then the value range of x+y is _ _ _ _ _ _ _.

5. The solution of equation = 3 is _ _ _ _ _ _ _ _.

6. the solution set of the inequality log (2- 1) log (2-2) < 2 is _ _ _ _ _ _ _.

A simple solution of 1: let sinx+cosx = t ∈ [-], then y =+t-, the symmetry axis t =- 1, when t =, y =+;

2 small problem: let x+ 1 = t (t ≥ 1), then f (t) = log [-(t- 1)+4], so the value range is (-∞, log4);

3 small problem: the known deformation is -=- 1, let b =, then b =- 1, b =-1+(n-1) =-n, so a =-.

4 small problem: let x+y = k, then x-2kx+ 1 = 0, △ = 4k-4 ≥ 0, so k≥ 1 or k≤-1;

5 small problem: let 3 = y, then 3 y+2 y- 1 = 0, and the solution is y =, so x =-1;

6 small problem: let log (2- 1) = y, then y (y+ 1) < 2, and the solution is -2.

Ⅱ. Presentation problem group:

Example 1. The real numbers X and Y satisfy 4x-S=x+y+4y = 5 (Formula ①). Let S=x+y and find the value of+. (93 National High School Mathematics League)

Analyze from S = X+Y to COS α+SIN α = 1, then carry out trigonometric substitution, and substitute into ① formula to find the values of S and S.

If the solution is substituted into ①, it is 4s-5s sin α cos α = 5.

Solve s =；;

∫- 1≤sin 2α≤ 1∴3≤8-5 sin 2α≤ 13∴≤

∴ +=+==

Through this solution, we can find the maximum and minimum values of S, and we can also find the inequality ||≤ 1 from the boundedness of sin2α= =. This method is a "bounded method" and is often used to find the range of function values.

Another solution is given by s = x+y, let x =+t, y =-t, t ∈ [-,],

Then xy =+/- is substituted into ① formula: 4s 5 = 5,

The square of the shift term is100t+39s-160s+100 = 0.

∴ 39s- 160s+ 100 ≤ 0 solution: ≤S≤

∴ +=+==

Note: The first solution to this problem belongs to the "triangular method of substitution", which mainly uses the relationship between the known condition S = X+Y and the trigonometric formula COS α+SIN α = 1 to associate, and finds that the algebraic problem is transformed into the trigonometric function range problem. The second solution belongs to the "average substitution method", which is mainly based on the equation S = X+Y. According to the idea of average substitution, let X =+T and Y =-T, which reduces the number of elements and is easy to solve. In addition, several methods are used to evaluate the domain: bounded method, inequality property method and separation parameter method.

Similar to "mean method of substitution", we also have a method of substitution, that is, when there are two variables x=a+b Y in the topic, x=a+b and Y = A-B can be set, which is called "sum and difference method of substitution", and method of substitution may simplify the algebraic expression. Let x = a+b, y = a-b, substitute ① to get 3a+ 13b = 5, and get a∈[0,], so s = (a-b)+(a+b) = 2 (a+b) =+a ∈.

Example 2. Delta ABC's three internal angles A, B and C satisfy: A+C = 2B,+=-,and find the value of cos. (1996 national management)

By analyzing the properties of "A+C = 2b" and "the sum of the internal angles of the triangle is equal to 180", we can get that "a+c = 120" is substituted into the mean value, then set, and then substituted to find cosα, which is cos.

The solution is obtained from A+C = 2b known in △ABC.

Let a+c = 120 and substitute it into the known equation:+=+=-2,

The solution is: cos α =, that is, cos =

Another solution comes from a+c = 2b, a+c = 120, and b = 60. So+=-

=-2, let =-+m, =-m,

Therefore, cosA= =, cosC= =, these two formulas are added and subtracted respectively:

cosA+cosC=2coscos=cos=，

cosA-cosC =-2 sin in =-sin =，

That is: sin =-, =-,substitute sin+cos = 1 to get: 3m- 16m- 12 = 0, solve m = 6, substitute cos = =.

Note: The two solutions to this problem are "A+C = 120" and "+=-2" respectively, and combined with the relationship between triangle angle and triangle formula, besides the known idea of mean replacement, you need to be quite skilled in the application of triangle formula. If the mean value is not expected to be substituted, it can be directly solved by trigonometry: A+C = 2b, and A+C = 120, and B = 60. So +=-=-2, that is, COSA+COSC =-2 COSA COSC, and the sum product is reciprocal:

2coscos =-[COS (a+c)+COS (a-c), that is, COS =-COS (a-c) =-(2cos- 1), and the sort is 4cos+2cos-3 = 0.

Solution: cos= =

y，，- x

Example 3. Let a>0 find the maximum and minimum values of f (x) = 2a (sinx+cosx)-sinx cosx-2a.

Let sinx+cosx = t, then t∈[-], from (sinx+cosx) =1+2 sinx cosx: sinx cosx =

∴f(x)= g(t)=-(t-2a)+(a & gt； 0)，t∈[-，]

When t =-, take the minimum value: -2a-2a-

When 2a≥, t =, and the maximum value is-2a+2a-;

When 0

The minimum value of ∴ f(x) is -2a-2a- and the maximum value is.

Note that this problem belongs to local method of substitution. Let sinx+cosx = t, grasp the internal relationship between sinx+cosx and sinx+cosx, and transform the range problem of trigonometric function into the range problem of quadratic function in closed interval, which makes it easy to solve. In the process of substitution, we must pay attention to the range of the new parameter (t∈[-,]) corresponding to sinx+cosx, otherwise an error will occur. The solution of this problem also includes the mathematical thinking method of classified discussion when it comes to problems with parameters, that is, the parameters are determined by the positional relationship between the axis of symmetry and the closed interval, and are discussed in two cases.

Generally speaking, when the known problem and the unknown problem contain the sum, difference and product of sinx and cosx, and the maximum and minimum values of a triangle are found, that is, the function is f (SINX COSX, sinxcsox), the substitution method of this argument is often adopted, which is transformed into the study of quadratic function or linear function in a closed interval.

Example 4. Let all real numbers x, inequality xlog+2xlog >; 0 is a constant, so find the range of A (National Management 1987).

What is the relationship between logarithm, logarithm and logarithm in analytic inequality? It is not difficult to find out the relevant deformation of logarithmic formula before implementing substitution method.

Solve log = t, then log = log = 3+log = 3-log = 3-t, log = 2 log =-2t,

After substitution, the original inequality is simplified to (3-t) x+2tx-2t >; 0 is true for all real numbers x, so:

The solution is ∴ t < 0 is logarithm.

0 & lt& lt 1, 0

The application of local substitution method has played a role in simplifying the complex and turning the difficult into the easy. Why do you think of changing people and how to set them up? The key is to find the relationship among log, log and log in the known inequality. When solving inequality problems, the "discriminant method" is used. In addition, this problem also requires very skilled logarithmic operation. Generally speaking, we can use local method of substitution to solve the inequality and equation between exponent and logarithm, or we can appropriately deform the given known conditions when replacing elements, find the relationship between them and implement the replacement, which is something we should pay attention to when thinking about the solution.

Example 5. Given = and+= (② formula), find the value of.

Solution = = k, then sin θ = kx, cos θ = ky, sin θ+cos θ = k (x+y) = 1, that is,+= = that is,+=

Let = t, then t+=, the solution is: t = 3 or ∴ = or.

Another solution is = = tgθ, divide both sides of equation ② at the same time, and then express it as a formula containing tgθ: 1+tgθ = = tgθ, let tgθ = t, then 3t- 10t+3 = 0,

∴ t = 3 or, the solution is = or.

Note: The first solution is equivalent substitution and substitution starting from =, which reduces the number of variables. The second solution, known as deformation =, is not difficult to find that the result is tgθ, and then substitute deformation. These two solutions require clever algebraic deformation. When solving higher-order equations, substitution method is used to reduce the number of equations.

Example 6. The real numbers x and y satisfy += 1, if x+y-k >; 0 is a constant, find the range of k.

By analyzing the known condition += 1, we can find that it is similar to A+B = 1, so we realize triangle replacement.

The solution is += 1, let = cos θ, = sin θ,

That is, substituting the inequality x+y-k >; 0 get:

3 cosθ+4 sinθ-k & gt； 0, that is, k < 3cosθ+4sinθ=5sin(θ+ψ).

So k < At -5, the inequality holds.

Note: In this question, the algebraic problem (or analytic geometry problem) is transformed into a trigonometric inequality problem with parameters, and then it is transformed into a trigonometric function range problem through "parameter separation", so as to find the parameter range. Generally speaking, when encountering algebraic expressions of equations similar to circles, ellipses and hyperbolas, or when solving problems related to circles, ellipses and hyperbolas, "triangular method of substitution" is often used.

y x x+y-k & gt； 0

K-plane region

Another way to solve this problem is to use the combination of numbers and shapes: in the plane rectangular coordinate system, the inequality ax+by+c > 0 (a > 0) indicates that the area is the part containing the positive direction of the X axis in the two parts of the plane divided by the straight line AX+BY+C = 0. The inequality problem is transformed into a graphic problem: the points on the ellipse always lie on the plane X+Y-K >; The area of 0. That is, when the straight line x+y-k = 0 is below the tangent of the lower part of the ellipse. When the line is tangent to the ellipse, the equations have a set of equal real number solutions, and after elimination, k =-3 can be obtained from △ = 0, so k < At -3, and the original inequality holds.

Ⅲ. Combined problem groups:

It is known that f (x) = lgx (x >; 0), then the value of f(4) is _ _ _ _.

A.2lg2 B. lg2 C. lg2 D. lg4

The monotonic increasing interval of the function y = (x+ 1)+2 is _ _ _ _.

A.[-2，+∞) B. [- 1，+∞) D. (-∞，+∞) C. (-∞，- 1]

Let the tolerance of arithmetic progression {a} be d = and S = 145, then the value of A+A+A+ ... +A is _ _ _.

A.85 B. 72.5 C. 60 D. 52.5

Given that x+4y = 4x, the value range of x+y is _ _ _ _ _ _ _ _ _ _ _.

Given a≥0, b≥0, a+b = 1, the value range of+is _ _ _ _ _ _ _.

The solution set of inequality > ax+ is (4, b), then a = _ _ _ _ _ _ _ and b = _ _ _ _ _ _.

The value range of the function y = 2x+ is _ _ _ _ _ _ _ _ _ _ _.

In the geometric series {a}, a+a+…+a = 2, a+a+…+a = 12, find a+a+…+a.

New York Times

What range does the real number m take? For any real number x, the inequality sinx+2mcosx+4m- 1

It is known that rectangular ABCD has vertex C (4,4) and point A on the curve X+Y = 2 (X >); 0, y>0), and AB and AD are always parallel to X axis and Y axis, so as to find the minimum area of rectangular ABCD.

Third, the undetermined coefficient method

The method of determining the functional relationship between variables, setting some unknown coefficients, and then determining these unknown coefficients according to given conditions is called undetermined coefficient method, and its theoretical basis is polynomial identity, that is, the necessary and sufficient conditions for using polynomial f(x)g(x) are: for any value of a, there is f (a) g (a); Or the coefficients of similar terms of two polynomials are correspondingly equal.

The key to solving the problem by undetermined coefficient method is to list the equality or equation correctly according to what is known. Using the undetermined coefficient method is to transform a mathematical problem with a certain form into a set of equations by introducing some undetermined coefficients. Judging whether a problem is solved by undetermined coefficient method mainly depends on whether the solved mathematical problem has a certain mathematical expression, and if so, it can be solved by undetermined coefficient method. For example, factorization, fractional decomposition, summation of series, finding functions, finding complex numbers, finding curve equations in analytic geometry, etc. These problems have clear mathematical expressions and can be solved by undetermined coefficient method.

Using the undetermined coefficient method, the basic steps to solve the problem are:

The first step is to determine the analytical formula of the undetermined coefficient problem;

Secondly, according to the condition of identity, a group of equations with undetermined coefficients are listed.

The third step is to solve the equations or eliminate the undetermined coefficients, so that the problem is solved.

How to list a set of equations with undetermined coefficients is mainly analyzed from the following aspects:

Use a column equation with equal corresponding coefficient;

Substitute the concept of identity into the normal equation by numerical value;

By defining its own attribute sequence equation;

The equations are sorted by geometric conditions.

For example, when solving the conic equation, we can use the undetermined coefficient method to solve the equation: first, set the form of the equation, which contains undetermined coefficients; Then the geometric conditions are transformed into equations or equations with unknown coefficients; Finally, the unknown coefficients are obtained by solving the obtained equations or equations, and the equations of conic curves are obtained by substituting the obtained coefficients into well-defined equation forms.

Ⅰ. Reproducible problem group:

Let f(x) =+m and the inverse function of f(x) = NX-5, then the values of m and n are _ _ _ _.

A.，-2 B. -，2 C，2 D. -，-2

Quadratic inequality ax+bx+2 >; The solution set of 0 is (-,), so the value of a+b is _ _ _ _.

A. 10 B. - 10 C

In the expansion of (1-x) (1+x), the coefficient of x is _ _ _ _ _.

A.- 297 BC-252 BC

Function y = a-bcos3x (b

The equation that the straight line L' is parallel to the straight line L: 2x+3y+5 = 0 and passes through the point A (1, -4) is _ _ _ _ _ _ _ _ _ _ _.

The hyperbola x-= 1 has an asymptote, and the equation of hyperbola passing through point (2,2) is _ _ _ _ _ _ _ _ _ _ _ _ _ _.

A simple solution to the 1 problem: find f (x) = 2x-2m from f (x) =+m, and the comparison coefficient is easy to find, so choose c;

2 Small problem: From the inequality solution set (-,), we can know that-is two of the equations AX+BX+2 = 0. Substitute two of them and list the equations about coefficients A and B. It is easy to get A+B, and choose D;

3 Small problem: the coefficient of X is composed of C and (-1) C, which are added to get the coefficient of X, and D is selected;

4 small question: list the equations of a and b from the known maximum and minimum values, find the values of a and b, and then substitute them to get the answer;

Question 5: Let the straight line L' equation 2x+3y+c = 0 and substitute it into point A (1, -4) to get c = 10, that is, 2x+3y+10 = 0;

6 Small problem: Let the hyperbolic equation x-= λ, and substitute it into point (2,2) to get λ = 3, and the equation -= 1.

Ⅱ. Presentation problem group:

It is known that the maximum value of the function y = is 7 and the minimum value is-1. Find this function.

The expression of analytical function is actually to determine the values of coefficients m and n; The known maximum and minimum values are actually the range of known functions, and the range of fractional functions whose numerator or denominator is a quadratic function is easily associated with "discrimination".

The transformation of the solution function is: (y-m) x-4x+(y-n) = 0, and x∈R is determined by the known y-m ≠ 0.

∴△ = (-4)-4 (y-m) (y-n) ≥ 0, that is, y-(m+n) y+(Mn- 12) ≤ 0 ①.

If the solution set of inequality ① is (-1, 7), then-1, 7 is the two roots of the equation Y-(m+n) Y+(Mn- 12) = 0.

Use two instead of get: get: or

∴ y= = or y=

This problem can also be solved by setting (y+ 1, 7) as (y+ 1) (y-7) ≤ 0, that is, y-6y-7 ≤ 0, and then comparing the coefficients with inequality ①, and solving m and n to get the function y.

Note: There are two coefficients m and n in the function formula to be determined. Firstly, the problem of function value domain is dealt with by "discriminant method", and a quadratic inequality about y with parameters m and n is obtained, and its solution set is known, so as to find parameters m and n. There are two methods to solve this problem. One is to consider two equations and list the equations after m and n are substituted. The second is to write inequalities from the known solution set, compare inequalities with parameters and list the equations of M and N to solve. This question requires a thorough understanding of the concept of the solution set of a quadratic inequality in one variable, and a "discriminant method" to find the function value domain: take Y as a parameter, and transform the function into a quadratic equation in one variable with parameter Y, so that it can be known that it has a solution. By using △≥0, the inequality about parameter y is established, and the range of solving y is range. The key to using "discriminant method" is whether the function can be transformed into a quadratic equation with one variable.

Example 2. Let the center of an ellipse be (2,-1), one of its focal points is perpendicular to the connecting line between the two ends of the short axis, and the distance from the focal point to the end point closer to the long axis is-,and then find the equation of the ellipse.

y B' x A F O' F' A' B

Analyze and solve the elliptic equation, and determine the values of geometric data A, B and C according to the given conditions. All problems are solved. After setting a, b and c, the known vertical relation is connected with Pythagorean theorem to establish an equation, and then the distance from the focus to the near end of the long axis is converted into the value of a-c, and the second equation is listed.

If the major axis 2a, minor axis 2b and focal length 2c of an ellipse are solved, | BF' | = a..

∴ Solution:

The elliptic equation is += 1.

You can also deduce the isosceles Rt△B'O'F' from its properties after deriving it from the vertical relationship, and then proceed as follows: It is easier to find the values of A and B.