The concept and nature of fractions are similar to fractions. For example, the denominator of a fraction cannot be zero, that is, a fraction is meaningful only if the denominator is not equal to zero. Like fractions, the numerator and denominator of fractions are multiplied (or divided) by the same algebraic expression that is not equal to zero, and the value of fractions remains unchanged. This property is the theoretical basis of general division and approximate division in fractional operation. In fractional operation, fractions are simplified mainly by reduction and general division, so as to evaluate fractions. In addition, the score should be flexibly deformed according to the specific characteristics of the score. This lecture mainly introduces the simplification and evaluation of fractions. Example 1 Simplified score: It is complicated to calculate the direct general score. First, each false fraction is converted into the sum of algebraic expression and true fraction. Simplification will be much simpler. = [(2a+1)-(a-3)-(3a+2)+(2a-2)] It shows that the key to this problem is to correctly write the false fraction as the sum of algebraic expression and true fraction. Example 2 When a=2, find the value of the fraction. Analyze and solve first, then evaluate. Just split it up. Scores can be divided into steps, and each step is only divided into two items on the left. Example 3 If abc= 1, we can separate the scores and simplify the evaluation, but it is more complicated. Here are a few simple solutions. The solution 1 is because abc= 1, so a, b and c are not zero. Solution 2 is because of ABC. C ≠ 0。 Example 4 Simplifying Fractions: It takes a lot of calculation to analyze and solve the three fractions together. The denominator of each fraction can be factorized first and then simplified. Explains a pair of opposites that cancel each other out. This simplification method is called "decomposition and elimination method" and is a common skill in fractional simplification. Example 5 simplifies the calculation (where A, B and C are two, obviously the denominator can be decomposed into (a-b)(a-c), and the numerator just adds up to (a-b)+(a-c), so there is the following solution. The solution shows that this example also adopts "division elimination", but the difference is that Example 6 shows that the score of x+y+z = 3a(a≦) is a complex number. If the condition is written as (x-a)+(y-a)+(z-a)=0, then the problem is only related to x-a, y-a, z-a, and method of substitution can be used to simplify the calculation. The solution is x-a=u, y-a=v, z-a = So u2+v2+w2≠0. From this example, we can see that method of substitution can reduce the number of letters and simplify the operation process. Example 7 Simplify factorial: deform appropriately, and then calculate and evaluate. (x-4) 2 = 3, that is, x2-8x+ 13 = 0. Protomolecule = (x4-8x3+13x2)+(2x3-16x2+26x)+(x2-8x+13)+10 = x2. +10 = 10, and the denominator of the original formula =(x2-8x+ 13)+2=2, indicating that the solution of this example adopts the whole substitution method, which is a special type of substitution elimination method. Proper application will greatly simplify the process of solving the problem.