First, the matching method
Matching method is a technique of directional deformation of mathematical formula (matching into a "complete square"), and the relationship between known and unknown is found through the formula, thus simplifying the complex. When making a formula, we need to make an appropriate prediction and use the skills of "splitting" and "adding", "matching" and "gathering" reasonably to complete the formula. Sometimes called "matching method".
The most common formula is identical deformation, so that the mathematical formula appears completely square. It is mainly suitable for the discussion and solution of known or unknown quadratic equations, quadratic inequalities, quadratic functions and quadratic algebra, or the translation transformation of quadratic curves without xy terms.
The most basic formula used in the matching method is the binomial complete square formula (A+B) = A+2AB+B, and various basic formula forms can be obtained by using this formula flexibly, such as:
a+b =(a+b)-2ab =(a-b)+2ab；
a+a b+b =(a+b)-ab =(a-b)+3ab =(a+)+(b)；
a+b+c+a b+ BC+ca =[(a+b)+(b+c)+(c+a)]
a+b+c =(a+b+c)-2(a b+BC+ca)=(a+b-c)-2(a b-BC-ca)=…
Combined with other mathematical knowledge and properties, there are some other formula forms, such as:
1+sin 2α= 1+2 sinαcosα=(sinα+cosα)；
x+=(x+)-2 =(x-)+2； ..... and so on.
Ⅰ. Reproducible problem group:
1. In positive geometric series {a}, a? a +2a？ a +a？ A =25, then A+A = _ _ _ _ _.
2. The equation x+y-4kx-2y+5k = 0 represents a circle if and only if it is _ _ _ _.
A.& ltk & lt 1 b . k & lt； Or k> 1 c.k ∈ r d.k = or k = 1.
3. Given that sin α+cos α = 1, the value of sin α+cos α is _ _ _ _ _.
A.1B.-1C.1or-1d.0.
4. The monotonic increasing interval of the function y = log (-2x+5x+3) is _ _ _ _ _.
A.(-∞，] B. [，+∞) C. (-，] D. [，3)
5. If the equation x +(a-2)x+a- 1=0 is known, then the point P(x, x) is on the circle x +y =4, and the real number A = _ _ _ _ _
A simple solution to the 1 problem: Using the property of geometric series a +a = A, the formula (a +a) after the left side of the equation is known is easy to find. The answer is: 5.
2 small problem: the standard equation form of formula rounding (X-A)+(Y-B) = R, and the solution R >；; 0 is enough, choose B.
3 Small problem: It is known that the equation is formulated as (sin α+cos α)-2 sinαcosα = 1, and then the square value of the formula is found, and then the root is found. Choose C.
Small problem: find the symmetry axis after the formula, and solve it by combining the definition domain with the monotonicity of logarithmic function and composite function. Choose D.
Question 5: Answer 3-.
Ⅱ. Presentation problem group:
Example 1. Given that the total area of a cuboid is 1 1 and the sum of the lengths of its 12 sides is 24, then the length of a diagonal of this cuboid is _ _ _ _ _.
A. May 6, 2 BC
Firstly, the analysis is converted into a mathematical expression: let the length, width and height of a cuboid be x, y, z, y and z respectively. Then, if you want to find the diagonal length, you can get it by combining it into two known formulas.
Let the length, width and height of a cuboid be x, y, z y and z respectively, and the total area of the cuboid is known as 1 1, and the sum of the lengths of its 12 sides is 24.
The diagonal length of the cuboid = = = 5.
So choose B.
Note that the key to solving this problem is to transform two known and one unknown into three mathematical expressions. Through the observation and analysis of three mathematical expressions, it is easy to find that the three mathematical expressions are connected by collocation, that is, the known and the unknown are connected, so as to solve them. This is also a problem-solving model that we use collocation method.
Example 2. Let the two real roots of the equation X+KX+2 = 0 be p and q, and if ()+() is less than or equal to 7, then the value range of the number k is realistic.
The two real roots of the equation x+kx+2 = 0 are p and q, which are obtained by Vieta theorem: p+q =-k, pq = 2,
()+() = = ≤ 7, and k ≤- or k≥
And ∵p and q are two real roots of the equation x+kx+2 = 0, ∴△ = k-8 ≥ 0, that is, k≥2 or k ≤-2.
On the whole, the range of K is: -≤k ≤- or ≤k≤
Pay attention to the quadratic equation with real coefficients, always consider the discriminant "δ" of the root first; When there are two known equations, Vieta's theorem can be used appropriately. In this topic, after Vieta's theorem obtains P+Q and pq, we observe the known inequality. Starting from its structural characteristics, we associate it with the formula and express it as the combination of P+Q and pq. If we don't discuss "△" in this question, the result will be wrong. Even though some questions may have the same result and the discussion of "△" is omitted, the answer is not rigorous and complete. We should pay special attention to this point.
Example 3. Let non-zero complex numbers a and b satisfy A+AB+B = 0, and find ()+().
The analysis can be associated with the known formula: if the deformation is () +()+ 1 = 0, then = ω (ω is the cubic imaginary root of 1); Or the formula is (a+b) = ab. And then substituting into the required formula to obtain the product.
The solution is transformed from A+AB+B = 0: ()+()+ 1 = 0.
Let ω =, then ω+ω+ 1 = 0, which means ω is the cubic imaginary root of 1, so: =, ω= = 1.
It is also transformed from A+AB+B = 0 to (a +ab+b =0 AB.
So () +() = ()+() = ()+() = ω+= 2.
Note that this problem simplifies the expression by formula; By skillfully using the cubic imaginary root of 1 and flexibly using the properties of ω, the higher power in the expression can be calculated. A series of transformation processes are flexible and require us to be good at association and expansion.
Another solution is changed from A+AB+B = 0 to: () +()+ 1 = 0. Solve =, convert it into triangular form, and then substitute it into the variable of the expression ()+() to complete the following operations. This method is only used to solve problems when ω is irrelevant.
If this problem does not think of the above-mentioned series of transformation processes, it can also be solved by A+AB+B = 0: A = B, directly substituted into the required expression, simplified by fraction, transformed into the triangular form of complex number, and the final calculation can be completed by using Dimover theorem.
Ⅲ. Combined problem groups:
1. The minimum value of the function y = (x-a)+(x-b) (a and b are constants) is _ _ _ _.
The minimum value of a.8b.c.d does not exist.
2.α and β are two real roots of the equation x-2ax+a+6 = 0, so the minimum value of (α- 1) +(β- 1) is _ _ _.
A.-b.8c. 18d. does not exist.
3. Given x and y∈R, x+3y- 1 = 0 is satisfied, and the function t = 2+8 has _ _ _ _.
A. Maximum value 2 b. Maximum value C. Minimum value 2 b. Minimum value
4. A focus of the ellipse x-2ax+3y+a-6 = 0 is on the straight line x+y+4 = 0, then a = _ _ _ _.
A.2 B.-6 C.-2 or -6 D.2 or 6
5. Simplification: the result of 2+ is _ _ _ _ _.
A.2 tin 4 boron 2 tin 4-4 cobalt 4 carbon
6. Let f and f be the two focal points of hyperbola -y = 1, point p is on hyperbola and ∠ f pf = 90, then the area of △F PF is _ _ _ _ _ _ _.
7. If x>- 1, then the minimum value of f(x)= x+2x+ is _ _ _ _ _ _.
8. known; 1，m∈R，x=log t+log s，y=log t+log s+m(log t+log s)，
① express y as a function y=f(x y = f(x), and find the domain of f(x);
② If the equation f (x) = 0 about x has only one real root, find the value range of m..