Inequality problems, a favor from the master 10

Solution: If the number of children is x, the number of apples can be (3x+8) according to the question.

Analysis: 5 per person and less than 3 last children. If there are 5 apples per person, then 5x apples are given to these 5 people, and because there are still 3 apples that have not been distributed, it means 5x<3x+8

Let's assume that the last child gets 3 apples, but the condition is that he gets less than 3. This yields: 5x+3>3x+8

From the above obtained set of two unary inequalities, we can get :

5x<3x+8； x<4

5x+3>3x+8； x>

So the set of solutions for this group of unary inequalities is and since x is an integer, we get x 3.

A: There are 3 children and 17 apples.

Analysis: Both people and apples should be positive integers, so the number of people can be set to x, the number of apples y, and finally the number of apples is less than 3, and there are only three cases, that is, 2, 1, or 0 left, which are discussed in 3 cases: three systems of equations to solve.

The solution is very simple, I won't do the math for you here, but it's important to check the result if it's not an integer!

Solution: Let there be x children.

0＜3x+8-5x＜3

0＜-2x+8＜3

8＜-2x＜-5

4 x Because x is an integer, x=3

So apples are 3 3 + 8 = 17

Untie; If there are x children, then there are (3x+8) apples

3x+8>5(x-1)

3x+8-5(x-1)<3

Solve x <

x>5 so x 6 A has 6 children and 26 apples

If you ask an algebra student to solve this problem, it should be simple: just set two unknowns. But it would be difficult to ask a student who had only studied arithmetic to solve the problem.

So sometimes the means or way of solving the problem determines whether the problem can be solved, at least affecting the efficiency of problem solving.

As for the simple solution, I won't go into details, but the solution of the brother on the first floor is very good.

Let the white ball b, the red ball h, by the title, bh, 2b+3h=60, launch 4h<60<5h, by h as an integer, so only 13,14, and 60-3h=2b is an even number, so h can only take 14, at this time b=9 so 9 white balls, 14 red balls.

The white and red balls each have x,y, (x,y are positive integers) xy,x3x<3y<6x

2x+3y=60

5x<60<8x

8<=x<12

From the above formulas, only x=9 and y=14 satisfy the condition that x, y are positive integers, so there are 9 white balls and 14 red balls.

Set x white balls and y red balls.

Then 2x+3y=60, so y=(60-2x) 3 x y, 2x y, so x y 2x

i.e. x (60-2x) 3 2x

3x＜60-3x＜6x

Get 20 3 x 10

So x=8 or x=9

When x=8, y=44 3 (undesirable, discarded) When x=9, y=14

So 9 white balls and 14 red balls.

White ball x, red ball y, get the inequality:

2x+3y=60;

x5x<60;

2x>y; 8x>60;

> ,y=14;

Solution 1: 10000 (1000-500)=20, indicating that if the number of households is 20, each household needs to pay an overall initial installation fee of 500 yuan.

So the number of households in the community is 20, that is, at least 21 households.

Solution 2: If the number of households in this community is at least x, then.

500x+10000<1000x

x+20<2x

There are at least 21 households in 20 communities.

If the number of households is a, then 10000 + 500a "1000a", calculate a>20

So at least 21 households.

The maximum value is negative.

That is, the image of this quadratic function is open downward.

And it has no intersection with the x-axis, that is, the discriminant formula is less than 0

So it's available. a<0△=16-4a(a-3)

4a²+12a+16

4(a²-3a-4)

4(a-4)(a+1)<0

Get A<-1 or A>4

To sum up a<-1

a<0 and a<-1 intersect to get a<-1a<0 and a>4 to intersect to get an empty set, so the range is a<-1

5 minutes wide oak = 1 12 hours (50-5) = 45 minutes = hours 3 * (1 12) = km km Buses should travel at least x km per hour ( x = 13 km Buses travel at least 13 km per hour.

Since the inequality is a loss homogeneous formula, it is advisable to let a+b+c=1, then the original formula is proved.

Root number (A (1-A)) + Root number (B (1-B)) + Root number (C (1-C)) > 2. The next uses the partial inequality to prove the binolet.

Observe the characteristics of the original formula first, and you can take it first.

Root number (a (1-a)) > 2a

Or. Root number (A(1-A))>B+C).

However, it seems that the former is easier.

The former can be transformed into 1 2> = root number (a(1-a)), that is, (a-1 2) 2>=0, which is obviously true.

In the same way, (b (1-b)) > 2b, (c (1-c)) > 2c, the above three formulas are added to obtain the root number (a (1-a)) + root number (b (1-b)) + root number (c (1-c)) > 2

The equal sign is only true when a=b=c=1 2, but at this time a+b+c=3 2 is not equal to 1

So the root number (a (1-a)) + the root number (b (1-b)) + the root number (c (1-c)) > 2

That is, the root number [a (b+c)] + the root number [b (a+c)] + the root number [c (a+b)] >2 is established.

The set price is $x.

If the original weight is A, then it is after loss.

So the arms cost a total of yuan.

And what you get from selling is x*

Therefore, if you do not lose money, you will be x*

A is about x

So x minimum is.

So at least meta.