People's education edition eighth grade first volume mathematics courseware
The first class is a comprehensive review.
I. Knowledge structure
Two. Summary of important knowledge and laws
(A) the concept
1, fraction: (A and B are algebraic expressions, and B≠0)
2. The simplest common denominator: the product of the highest power of all factors of each denominator.
3. Fractional equation: an equation with an unknown number in the denominator.
(2) Nature
1, the basic property of the fraction: (m is an algebraic expression not equal to zero)
2, the nature of power:
Zero exponential power: = 1(a ≠0)
Negative integer exponential power: (a≠0, n is a positive integer)
Scientific notation: a ×, 1≤| a |< < 10, where n is an integer.
(3) Fractional algorithm
Fractional multiplication: Multiply the numerator and denominator separately, that is
Fractional division: after the numerator and denominator of division are reversed, multiply by the divisor, that is
Addition and subtraction of fractions: (1) Addition and subtraction of fractions with the same denominator:;
(2) Addition and subtraction of fractions with different denominators:
Fractional power: (b≠0) Fractional root: (a≥0, b>0)
(4) solution of fractional equation
1, solution idea: transform fractional equation into integral equation.
2. Conversion method: denominator (especially substitution method).
3. The key to transformation: correctly find the simplest common denominator.
4. Note: Pay attention to root measurement.
Third, the inspiration of learning methods
1, when two integers are not divisible, a score appears; Similarly, when two fractions are not divisible, a fraction appears. Therefore, the division of algebraic expressions is the basis of introducing the concept of fraction.
2. The basic properties and operations of fractions are similar to those of fractions. Therefore, in the process of learning, we should pay attention to constant analogy with scores and deepen our understanding of new knowledge.
3. The idea of solving the fractional equation is to remove the denominator containing the unknown, so as to transform the fractional equation into an integral equation to solve it. At this time, there may be root growth and it must be tested. When studying, we should understand the reasons of root growth, realize the necessity of inspection and carry out inspection.
4. Because zero exponential power and negative integer exponential power are introduced, numbers with smaller absolute values can also be expressed by scientific notation.
Fourth, assign homework: review the problems in chapter 16 of the textbook.
Special lecture in the second class
First, the common skills in decimal operation
The operation of fraction is based on the concept, basic properties and operation rules of fraction, among which the addition and subtraction of fraction is difficult. The key to solve this problem is to divide the score appropriately according to the characteristics of the topic, and calculate it with algebraic deformation and factorization as tools. Fractional operation not only highlights the basic knowledge and skills of algebraic operation and transformation, but also pays attention to the thinking method of mathematics, which is one of the key contents of previous exams. If we can choose the solution flexibly according to the characteristics, we will get twice the result with half the effort.
1, approximate evaluation: when denominator or numerator is polynomial, factorize numerator and denominator first, and then approximate evaluation.
Calculation:
Solution: Original formula =
2. Step by step general division, step by step calculation: explain the solutions to the following problems. This problem adopts "gradual general division method". First, divide the first two scores, and then divide the result with the following scores, simplifying the complex. If all of them are dropped at one time, the calculation will be very large. When solving problems, we should not only see local characteristics, but also consider them comprehensively.
Calculation:
Solution: Original formula =
3, reasonable collocation, group sharing: group sharing can reduce the difficulty, see the following question.
Given x= 1+, then = _ _ _ _ _ _ _ _ _.
Analysis: separate the first item from the third item, then use the second item to calculate, and finally substitute it for evaluation.
Second, the common skills of score evaluation
Score evaluation frequently appears in the senior high school entrance examination, and the method is flexible. Sometimes, it is difficult to simplify and deform conditions or algebraic expressions. When the numerator and denominator of contemporary number expressions are reversed, it is easy to deform. This kind of problem is usually evaluated by reciprocal method (the numerator and denominator are reversed), such as 1.
For example, the value of 1 is known.
Solution: that is, ∴x≠0.
∴ ,∴ = 。
2. Flexible application of formula deformation evaluation: If the formula can be skillfully deformed, it will bring great convenience to solving problems, as shown in Example 2.
Example 2: Given x2-5x+ 1=0, find the value of.
Solution: From x2-5x+ 1=0, x≠0 is known, from which it is obtained.
∴
3. Set K evaluation method (also called parameter method): When the known conditions appear in the form of simultaneous equations, set K method is easier to solve problems, as shown in Example 3.
Example 3: The value of: is known.
Solution: Let =k, ∴b+c=ak, c+a=bk, a+b=ck.
∴b+c+c+a+a+b=ak+bk+ck，
∴2(a+b+c)= k (a+b+c)，(a+b+c)(2-k) =0
That is, k=2 or a+b+c=0 and substitution = K.
∴ Original formula =. That is, the original formula = or the original formula =- 1
4. Whole replacement method: When calculating the algebraic evaluation problem, sometimes the whole replacement method can be used, that is, the conditional equation (or deformed conditional expression) is replaced and evaluated as a whole, as shown in Examples 4 and 5.
Example 4. Given the values of,, and.
Solution: ∵,,,
∴ ,∴ = 。
∴ 。
Example 5: Given a+b=-8 and ab=6, simplify _ _ _ _ _ _ _ _.
Solution: ∫a+b =-8, ab=6, ∴ A.
∴ Original formula =
Third, homework
Review the questions in chapter 15 of the textbook.