Kneel down and ask for 80 application problems to solve inequalities!! With answers!! Thank you! I m

1. Multiple choice questions.

1. If a, b are arbitrary real numbers, and a b, then (

a）a2＞b2 （b） ＜1 （c）lg(a-b)＞0 （d）( a＜( b

2. The following inequalities are true (

a）lgx+logx10≥2(x＞1) （b） +a≥2 (a 0)

c） ＜a＞b) （d）a ≥a (t＞0,a＞0,a 1)

3. Knowing that a >0, b >0 and a +b=1, then the minimum value of ( is (

a)6 (b) 7 (c) 8 (d) 9

4. The following inequalities have been given: (1)x3+ 3 >2x(x r); 2) a5+b5> a3b2+a2b3(a ,b∈r);

3) a2+b2 2(a b 1), where the correct number is (

a) 0 (b) 1 (c) 2 (d) 3.

5. f(n) = n , n) = , g(n) = n , n n, then (

a) f(n)1.

10 Solving inequalities

Inequality practice answers.

1. DADCB ddDAB BCBAB

II, 1, (a1 a2 ....)an x 1 or x

x＜log23 9、-3＜x≤2

10, x 0 or 1 x 4

Three, 1, [1] (1, )2, (1,0) (0,3) 3, (2) (3, 4, (0,3), 7, 8, 2 m 0

9. Solution: (i) When a>1, the original inequality is equivalent to the group of inequalities:

Solution x>2a-1

ii) When 01, the set of solutions of the inequality is;

When 010, the original is not equivalent to the inequality group (1) or (2).

From (1) , from (2) to get x 3, so the set of solutions of the original inequality is.

1.During blasting construction, the speed at which the fuse burns is 5m s, so how long does the fuse need to be at least in order for the ignition soldier to run to a safe area 100m away during construction?

2.If an engineering team stipulates that 300 earthworks should be completed in 6 days, and 60 earthworks are completed on the first day, and now it will be completed at least two days earlier than originally planned.

3.It is known that Li Hong is 3 years older than Wang Li, and it is known that the sum of the ages of Li Hong and Wang Li is greater than 30 and less than 33, so Li Hong's age is asked.

4.A worker plans to process 408 parts in 15 days, and 24 parts per day for the first three days.

5.Wang Kai's home is kilometers to the school, and now he needs to walk this section in 18 minutes. It is known that Wang Kai walks at a speed of 90 meters and runs at a speed of 210 meters.

6.An engineering team plans to repair 6km of road in 10 days, and after the construction is completed 2 days before construction, the plan changes, and the road repair task is ready to be completed 2 days in advance.

Answer: 1Solution: Set the fuse XCM

x÷x≤16 2.After that, at least x square soil should be completed every day.

6-2)x≥300-60

x≥60 3.Let's take Li Hong's age to be x years old.

30≮x+(x-3)≮33

x 18x must be an integer x take 17

4.After setting, at least x will be processed per day.

15-3)x≥408-24×3

x≥96 5.Set x points for running and 18-x points for walking.

90(18-x)+210x≥2100

x≥4 6.Solution: After setting up the road to be repaired every day x kilometers, then.

10-2-2)x≥

6x x Answer: In the future, at least kilometers of roads will be built every day.

If A and B start from a certain place at the same time and travel in the same direction, and B walks at a speed of 5 kilometers per hour, and sets off 2 hours before A, if A rides a bicycle to catch up with B within 40 minutes (including 40 minutes), what is the minimum speed of A riding a bicycle?

Solution: Set the speed of the armor to be x kilometers per hour. 40 minutes 2 to 3 hours.

2/3 x－5×2/3≥5×2

2/3x≥40/3

x 20A: A rides at a speed of at least 20 km/h.

Fourth, a city produces an average of 700 tons of domestic garbage per day, all of which are processed by two garbage plants A and B, and it is known that Plant A processes 55 tons of garbage per hour, which costs 550 yuan; Plant B processes 45 tons of garbage per hour, which costs 495 yuan. If the cost of garbage disposal in the city is not more than 7,370 yuan per day, how many hours does Factory A need to dispose of garbage at least every day?

Solution: The garbage should be disposed of for at least a hour in the first place.

550a+（700-55a）÷45×495≤7370550a+（700-55a）×11≤7370550a+7700-605a≤7370

330≤55a

A 6A should dispose of garbage for at least 6 hours.

5. The school allocates a number of dormitories to the female students of the first class of the seventh grade, and it is known that there are less than 35 female students in the class, and if there are 5 students in each room, the remaining 5 people have nowhere to live; If there are 8 people in each room, one room will be vacant, and one room will not be satisfied. How many dormitories are there and how many girls are there?

Solution: If there is room A in the dormitory, the number of female students is 5A+5.

According to the title. a>0(1)

0<5a+5<35（2）

0<5a+5-[8（a-2）]<8(3)

Obtained from (2).

5<5a<30

From (1) 3a+8-5a+5<3

2a>10

a>5 by (2).

3a+8-5a+5>0

2a<13

a: "Then the value range of a is 5, then a = 6

There are 6 students and there are 3 6 + 8 = 26 books.

That's part of it, if you think it's okay.

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（2009?Chengdu) a university graduate in response to the country's call for "independent entrepreneurship", invested in the opening of a decoration store The store purchased a new jewelry on the market this year for 30 days of trial sales, the purchase of ** is 20 yuan After the end of the sale, it is learned that there is the following relationship between the daily sales volume p (pieces) and the sales time x (days): p = -2 x + 80 (1 x 30, and x is an integer); It is also known that there is the following relationship between the sales of **Q1 (yuan) and the sales time x (days) in the first 20 days:

Q1 = x+30 (1 x 20, and x is an integer), and the sales **Q2 (yuan) and the sales time x (days) in the next 10 days have the following relationship: Q2=45 (21 x 30, and x is an integer).

1) Try to write the functional relationship between the daily sales profit r1 (yuan) in the first 20 days and the daily sales profit r2 (yuan) in the next 10 days and the sales time x (days) respectively;

2) In this 30-day trial sale, which day has the largest daily sales profit? And find this maximum profit

Note: Sales profit = sales revenue - purchase cost Test point: Application of quadratic functions Special topic:

Application Problem Analysis: (1) Use the basic equivalence relationship in marketing problems: sales profit = daily sales volume sales profit of one piece Sales profit of one piece sales profit = sales price of one piece - purchase price of one piece, and establish a functional relationship;

2) Analyze the categories of the function relationship and the range of independent variables to find the maximum value; where r1 is a quadratic function and r2 is a quadratic function Answer: Solution: (1) According to the meaning of the question, obtain.

r1=p(q1-20)=(-2x+80)[(x+30)-20],-x2+20x+800 (1 x 20, and x is an integer), r2=p(q2-20)=(-2x+80)(45-20), -50x+2000(21 30, and x is an integer);

2) When 1 x 20 and x is an integer, r1=-(x-10)2+900, when x=10, the maximum value of r1 is 900, when 21 x 30 and x is an integer, r2=-50x+2000, -50 0, r2 decreases with the increase of x, when x=21, the maximum value of r2 is 950,950 900, when x=21 is on the 21st day, the daily sales profit is the largest, and the maximum value is 950 yuan Comments: This question needs to be read repeatedly to understand the meaning of the question, According to the basic equivalence relationship in the marketing problem, the function relationship is established, the segmented functions are listed according to the time period, and then the maximum values of the two functions are obtained by combining the value range of the independent variables, and the conclusions are drawn

First of all, solve the problem, break down the problem, two main points, one is to sit full, and the other is not to exceed the total cost of not more than 5000. (There is also a small trap in it: the four-seater car is 60 yuan each, and the eleven cars are 10 yuan for nobody, which is 110 yuan each).

The following solutions can be made from the above two points.

Set up to rent four-seater car x car, rent eleven car y.

11x+4y=70

110x+60y<=5000-60*70

Solving equations (methods that don't say the answers).

From the first equation, we can see that x is an even number and x<7;

Then x, y can be taken as 2 and 12 respectively 6， 1；

Substituting the second equation gives x=6,y=1;for the answer.

The range of the number of fertilizer bags produced is x

In 2005, the production volume of artificial labor was x<200*2100 4=105000 bags, and the production volume of raw materials in 2005 was x (600+1200)*1000 20=90000 bags.

In 2005, it is expected to sell 80,000 bags of fertilizer.

Based on the above analysis.

90000≥x≥80000

The maximum number of working hours of a worker is 200 2100 4 = 105000

The amount of raw materials that can be produced is 600 1200 1000 20=90000

Since the saleable volume is 80000 90000 105000, the number of bags produced is 80000 90000

1.Set A and B each A and B bottles.

6a+9b=780, a+b=100, get a=40, b=602If A x bottle, then B 2x bottle.

6x+18x≤1200

then x 50

1: From the meaning of the title, the stool filial piety + 2(20-x-y) = 42 is simplified: y = 20 - 2x ( 9 x jujube such as 2 , and x is an integer).

2：w = × 6x + 8y + 2 × 5 × 20 - x - y)

Y = 20 - 2x (w) = 336 - 9 x 2 (and x is an integer).

So when x = 2, i.e. when 2 cars are loaded with eggplant, the profit is maximum.

Released rented x cars.

36x-42(x-2)>30

36x-42(x-2)<42

Figure it out yourself.

Set, uproar. 36-seater good oak x car.

x-1)42-36x>30

6x-42>30

x> friends next to the reeds 16