(A) algebraic substitution method
Example solution equation -= 1
Solution: Let =t (t0)
Then = 1+t
So there are:
(1)-(2): t = 2 instead of (2):
2x2-3x-2 = 0: x 1 = 2，x2 =-
It is proved that x 1 = 2 and x2 =- are the solutions of the original equation.
Example 2 Verification: ()
Proof: Let y = then: x2+2 = y2+ 1.
So the original formula =
therefore
Summary: Example 1 Summary: By changing the elements, the complex operation of twice square in the conventional solution is avoided, which makes the problem easier to solve. This is called algebraic method of substitution. Example 2 makes the problem clearer by changing the elements. Then prove the inequality with the average value.
Example 3 Find the range of function y = sinxcosx+sinx+cosx.
Solution: let t = sinx+cosx = sin(x+)
Then t[]
sinx cosx =[(sinx+cosx)2- 1]=(T2- 1)。
So y = (t2-1)+t = (t+1) 2-1.
When t =-1, ymin =-1.
When t =, ymax =+
Therefore, the range of the function is [- 1,+].
(2) Constant substitution method
Example 4 shows that f(x) = 2x5+3x3-x2-4x+ 12. Find the value of f (1-).
Solution: Let 1-= x and then x2+2x- 1 = 0.
∫2x 5+3x 3-x2-4x+ 12 =(2x 3-4x 2+ 13x-3 1)(x2+2x- 1)+
7 1x- 19
= 7 1x- 19
∴f( 1-)= 7 1( 1-)- 19
= 52-7 1
Summary: Using constant substitution method to construct zero factor greatly reduces the amount of calculation. It fully embodies the subtle role of constant substitution method in solving problems.
Problem promotion:
Example 5 Given f(x-3) = 2x2+5x-6, find the analytical formula of f(x).
Solution: let x-3 = t then x = t+3.
Substituting x = t+3 into f(x-3) = 2x2+5x-6, we get:
f(t) = 2(t+3)2+5(t+3)-6
= 2t2+ 17t+27
So f(x) = 2x2+ 17t+27.
Summary: Constant substitution method is a common method to find the resolution function.
(3) Proportional substitution method
Example 6 If = = verification:
sin 2(α-β)+sin 2(β-γ)+sin 2(γ-α)= 0
Proof: Let = = =
Then x = artan (θ+α) y = artan (θ+β) z = artan (θ+γ).
sin 2(α-β)=〔cos 2(θ+β)-cos 2(θ+α)〕
sin 2(β-γ)=〔cos 2(θ+γ)-cos 2(θ+β)〕
sin 2(γ-α)=〔cos 2(θ+α)-cos 2(θ+γ)〕
Add the above three formulas to get:
sin 2(α-β)+sin 2(β-γ)+sin 2(γ-α)= 0
Summary: Pay attention to the structural characteristics of the question type, which is similar to the proportional formula. With proper substitution and triangular operation, the problem becomes simple and easier to solve.
(4) Standard substitution method.
Example 7 Let a 1, a2, a3, …, a2004 all be real numbers.
If a1+a2+a3+…+a2004 = 2004 … (1)
…… (2)
Validation: =2004
Proof: let A 1 = 1+M 1, A2 = 1+M2, A3 = 1+M3, …, a2004= 1+m2004.
According to the formula (1):
m 1+m2+m3+…+m2004=0 …… (3)
It can be obtained from formula (2)
( 1+m 1)2+( 1+m2)2+( 1+m3)2+…+( 1+m 2004)2 = 2004
Expand it and simplify it with (3):
=0
Therefore:
m 1=m2=m3=m2004=0
Namely:
a 1=a2=a3=…=a2004= 1
So:
Summary: In the example, the standard quantity is 1, a 1, a2, A3 ... A2004 are all represented by 1, and the auxiliary quantity is m 1, m2, m3, ... m2004. This method is called "standard substitution method".
(5) Triangle replacement method
Example 8( 1) knows that X >: 0, y >；; 0, find the minimum value of x+y.
(2) Solving inequalities:
Solution: (1) Let = cos2θ, sin2θ (0
So x+y=
= 10+tan 2θ+9 cot 2θ≥ 10+2 tanθ3 cotθ= 16
Therefore:
When tanθ=3cotθ, that is, X = 4 and Y = 12.
(x+y)min= 16
(2) Let x=2sinθ (-)
Then the inequality is: 2cosθ≥2sinθ.
Solution:-
So that -2≤2 sinθ≤ 1, that is, -2≤x≤ 1.
Note: If the value range of variable X can be converted into:-1≤x≤ 1 or-1
Example 9: 1 ≤ x2+y2 ≤ 2.
Verification: ≤ x2-xy+y2 ≤ 3
It is proved that ∵ 1 ≤ x2+y2 ≤ 2 can be set as x = rcos θ and y = rsin θ.
Where 0 ≤θ ≤2
∴ x2-xy+y2 = r2-r2 symplectic θcosθ= R2( 1- symplectic 2θ)
∫≤ 1-sin 2θ≤
∫R2≤R2( 1-sin 2θ)≤R2
And R2≥R2≤3.
∴ ≤ x2-xy+y2 ≤ 3
Summary: 1. For some inequalities, the problem is more obvious through proper triangle substitution.
2. Triangle substitution method can be used for all conditions, such as x2+y2 = 1 or x2+y2 ≤ r2. However, we should pay attention to the restrictive conditions of R and θ after substitution, such as the range of R in Example 9.
Popularization of the problem:
Example 10 In △ABC, A, B and C are opposite sides of angles A, B and C respectively. If c2=a2+b2, then △ABC is a right triangle. Now please study: If an+BN = CN (n > 2 and n∈N) Why is △ABC a triangle? Why?
Solution: ∫an+bn = cn Therefore, let an = cncos2θ bn = cnsin2θ (0
Then: a2=c2 b2=c2.
∴a2+b2= C2(+)& gt； c2()=c2
By cosθ= >0 means ∠C is an acute angle and c is the largest side.
So △ABC is an acute triangle.
(6) Incremental substitution method
Example 1 1 Verification: For any real number A >；; 1, b> 1 has inequalities.
Proof: let a= 1+x, b= 1+y, x, y∈R then
=
Take the equal sign if and only if x= y = 1, that is, a = b =2.
The solution to this problem is incremental substitution method. The so-called incremental substitution method is to replace m+t with the related variable x, where m is an appropriate constant, so strictly speaking, it is not necessarily incremental; In addition, this substitution is linear in nature, so the basic substitution of the above example 1 1 zhong 1-2y=t is also linear or incremental. Another example is:
Example 12 Find the maximum and minimum values of the function f(x)=+.
Solution: by
The solution is 4≤x≤5, that is, the domain of the function is: 4≤x≤5,
So x is a variable between 4 and 5.
Therefore, let x = 4+sin2θ (0≤θ≤), then
f(x) = sinθ+cosθ = 2sin(θ+)
When θ =, f(x) takes the maximum value of 2;
When θ =, f(x) takes the minimum value 1.
Summary: This case is both a triangular method of substitution and an incremental method of substitution. By transforming elements into triangular knowledge, the problem is solved skillfully.
Popularization of the problem:
Example 13 It is known that real numbers a 1, a2, a3, …, a8 satisfy a 1+a2+a3+…+a8=20.
a 1a2a3…a8= 12
Verification: at least one of a 1, a2, a3, …, a8 is less than 1.
Prove: (reduction to absurdity)
Let a 1, a2, a3, …, a8 be not less than 1.
Let A 1 = 1+T 1, A2 = 1+T2, A3 = 1+T3, …, A8 = 1+T8,
t 1，t2，t3，…，t8≥0
And t1+T2+T3+…+t8 = a1+a2+a3+…+A8-8 =12.
a 1 * a2 * a3 *……* A8 =( 1+t 1)( 1+T2)( 1+T3)……( 1+t8)
= 1+(t 1+T2+T3+…+t8)+…≥ 13
This is in contradiction with Yizhi a 1a2a3…a8= 12.
(7) Parameter substitution method
Example 14 It is known that x2+4y2+8x4+7=0. Find the minimum value of x2+y2 and the corresponding values of x and y. ..
Solution: From x2+4y2+8x4+7=0, we get: (x+4)2+4y2=9.
Re-deformation is:
Let (θ is the parameter 0≤θ≤2π)
Then:
x2+y2= 16-24cos
Because: cos θ
Therefore, when cosθ = 1, (x2+y2)min= 1 and x =-1 y = 0.
When cosθ =-1, (x2+y2)max= 49, x = -7 y = 0.
Summary: For the bivariate quadratic equation with conic condition, the maximum of structural formula F(x, y) of two variables x, y can be solved by parameter substitution method.
(8) Parameter variable substitution method
Example 15 calculation
Solution: let z = then z17 =-1z34 =1.
Similarly:
Original formula =
=
=
Example 16 calculated value
Solution: let x = > 0-( 1)
For the square on both sides of (1):
x2==
Solve equation 2x2+5x-3=0 again, and find the positive root to get:
X=, solved.
Example 17 It is known that ellipse = 1, fixed point A( 1, 1), if the chord PQ of point A is located.
Linear equation.
Solution: ∫ Point A( 1, 1) is the midpoint of the chord PQ.
Therefore, you can set p (1-m, 1-n) and q (1+m, 1+n).
∵ points p and q are on the ellipse = 1,
∴
( 1)-(2) -4m- 16n = 0。
y
p
that is
A
x
Q
o
∴ The slope of the straight line PQ is
kPQ = -，
Therefore, the linear equation of PQ is:
y- 1 = -(x- 1)，
Namely: x+4y-5 = 0
Summary: Setting variables to solve problems is also the application of method of substitution Thought. This problem adds an unknown number, skillfully combines the relationship between the endpoint coordinates and the midpoint coordinates of the line segment, making the problem clear and the process simple, which is the best emotional boundary for changing elements.
(9) Diversified alternative methods
Example 18 if a+b+c= 1, verification: a2+b2+c2≥
Proof: Let a=+α b=+β c=+γ.
Then it is: α+β+γ=0.
A2+b2+c2=(+α)2 +(+β)2 +(+γ)2。
= +2(α+β+γ)+(α2+β2+γ2)≥
Example 19 known
Verification:
Proof: order
The theme is set to:
From (2):
From ( 1): () 2 = 1。
Namely:
∴
Namely:
Summary: Because the quality and quantity of things are determined by many factors, if we change each of them, we may have new ideas. When solving mathematical problems, using "multiple substitution method" can simplify the problems and make them easier to solve.
To sum up, method of substitution plays an important role in solving mathematical problems. Summarizing the rules and skills of solving problems and strengthening thinking training will be very beneficial to improve students' ability to analyze and solve problems. It can also improve students' quality in an all-round way and cultivate and improve their creative ability. Therefore, it is more necessary for us to re-understand mathematical methods and improve teaching quality in an all-round way.