Analysis and Solution This is a decimal addition problem. If it is troublesome to add from left to right, it is observed that the sum of 3. 17+5.83, 2.74+0.26, 6.3+4.7 in the formula can all add up to an integer. So we can use additive commutative law and the law of association to calculate.

The original formula = (3.17+5.83)+(2.74+0.26)+(6.3+4.7)+5.29.

=9+3+ 1 1+5.29

=28.29

Learn and practice.

Calculate 6.11+9.22+8.33+7.44+5.55+4.56+3.67+2.78+1.89.

Example 2 calculates the following problem:

( 1)9.26-4.38-2.62

(2)9.26-(4.38+2.26)

(3)9.26-(4.38-2.74)

When analyzing and solving the problem of calculating the mixed operation formula of decimal addition and subtraction, according to the characteristics of data, the requirement of "rounding up" is met by adding brackets and removing brackets, which makes the calculation simple.

(1) Original formula =9.26-(4.38+2.62)=9.26-7=2.26.

(2) The original formula = 9.26-2.26-4.38=7-4.38=2.62.

(3) The original formula = (9.26+2.74)-4.38 =12-4.38 = 7.62.

Practice calculation while learning.

( 1)4.75-9.64+8.25- 1.36

(2) 14.529+(2.47 1-3)

(3)38.68-(4.7-2.32)

(4)7.93+(2.8- 1.93)

Example 3 calculates the following problem.

( 1)8×25× 1.25×0.04

(2)36÷ 12.5

(3)0.25× 1.25×32

Analyzing and solving these three problems are all mixed calculation problems of integer and decimal multiplication and division, which can be calculated by using the multiplication law and the invariance of quotient.

(1) The original formula = (8×1.25 )× (0.04× 25) =10×1=10.

(2) The original formula = (3600× 8) ÷ (12.5× 8) = 28800 ÷100 = 288.

Or the original formula = 36×100 ÷12.5 = 36× (100 ÷12.5) = 36× 8 = 288.

(3) The original formula = 0.25×1.25× (4× 8) = (4× 0.25 )× (1.25× 8) =10.

Practice calculation while learning.

( 1)64× 12.5×0.25×0.05

(2)27÷0.25

(3) 12.5×0.76×0.4×8×2.5

Example 4 calculates 0.1+0.2+0.3+…+0.9+0.10+0.12+…+0.98+0.99.

Analysis and solution: It is observed that this string number is not arithmetic progression, but consists of 0. 1 to 0.9 and 0. 10 to 0.99, which are arithmetic progression. So the sum of these two parts can be summed in groups and then summed.

The original formula = (0.1+0.9) × 9 ÷ 2+(0.10+0.99) × 90 ÷ 2.

=4.5+49.05

=53.55

Practice calculation1.1+3.3+5.5+7.7+9.9+113.15.

Example 5 Calculate the following problems

( 1) 7.24×0. 1+5×7.24+4.9×7.24

(2) 1.25×67.875+ 125×6.7875+ 1.25×53.375

(3)7.5×45+ 17×2.5

The multiplication and division method of analyzing and solving integers is not only applicable to integers, but also to fractional elementary arithmetic.

In (1), * * * has three products, and the factor of each product is 7.24, so it can be calculated by multiplication and division.

The original formula = 7.24× (0.1+5+4.9) = 7.24×10 = 72.4.

(2) The simple features are not obvious at first glance, but after careful observation, we can find that if 125×6.7875 is converted into 1.25×678.75 (think about it, why? ) All three products have a factor of 1.25, so it is very simple to calculate by multiplication and division.

The original formula =1.25× 67.875+1.25× 678.75+1.25× 53.375.

= 1.25×(67.875+678.75+53.375)

= 1.25×800

= 1000

(3) Since 45= 17+28, 7.5×45 can be converted into 7.5×( 17+28), the calculation can be simplified by using the algorithm.

The original formula = 7.5× (17+28)+17× 2.5 = 7.5×17+7.5× 28+17× 2.5.

= 17×(7.5+2.5)+7.5×4×7= 170+2 10=380

Think about it: What other factors can be divided to make the calculation simple?

Learn, practice and calculate by simple methods.

( 1)383.75×7.9+79×6 1.625

(2)9.99×0.7+ 1. 1 1×2.7

(3)6.25×0. 16+264×0.0625+5.2×6.25+0.625×20

Related links

Using the learned algorithm, operation properties, the changing law of differential product quotient, the summation formula of the sequence to be delayed, etc. , which can make some decimal calculations simple and convenient. It is worth noting that some simple decimal calculations with insignificant characteristics can become simple and flexible only after reasonable deformation. For example, in the last two cases in Example 5, two points should be reminded during deformation: (1) The hidden simple calculation features should be exposed after deformation; (2) The deformation size cannot be changed.

Extracurricular expansion uses a simple method to calculate the following questions.

( 1)34.5 8.23-34.5+2.77 34.5

(2)6.25 0. 16+264 0.0625+5.2 6.25+0.625 20

(3)0.035 935+0.035+3 0.035+0.07 6 1 0.5

(4) 19.98 37- 199.8 1.9+ 1998 0.82

(5) 1-0. 1-0.0 1-0.00 1-0.000 1-……-0.00000000 1

Walk into the topic of competition

1,12.5× 69+53× 3.1+72× 3.1(Mathematics Competition for Primary School Students in Wuyuan County, Jiangxi Province in 2003)

2. 1. 1+3.3+5.5+7.7+9.9+ 1 1+ 13. 15.66.

3. 0.79 × 0.46+7.9 × 0.24+11.4 × 0.079 (the first imagination cup trial in 2003)

4, 7.5×23+3 1×2.5(2002 Kaiping Grade 5 Competition Question)