A junior high school inequality problem, junior high school math inequalities

As can be seen from the title, when the printed material is less than 1,000 copies, the cost of the two printing houses is the same;

When the printed materials are more than 1000 copies and less than 2000 copies, it will be more preferential to Factory A;

When it is necessary to print x copies of promotional materials (x>= 2000), the printing fee of both factories is the same.

There are 500+ at this point

Solution x=3000

Answer: When the printed materials are less than 1000 copies or equal to 3000 copies, the cost to the two factories is the same, when the printed materials are greater than 1000 copies and less than 3000 copies, to factory A is more preferential, when the printed materials are greater than 3000 copies, to factory B with discounts.

When the number of copies is greater than 3000, factory B is more favorable; When the number of copies is less than 3000, factory A will be discounted.

Set up printing x copies, x>2000.

A printing factory asking price: y1 = 500+

B printing factory asking price: y2=500+

y1-y2=, because x>2000, so y1-y2>0, so factory B should be chosen.

You first set to print x copies, then calculate x cents, and finally compare them.

Because it is required that the area of planting turf and planting trees is not less than 10 acres, the area of planting turf is x 10, and the area of planting trees is 30-x 10, that is, x 20, so 10 x 20

And the cost of greening = 8000x + 12000 (30-x) = 360000-4000x.

The cost of greening decreases as x increases.

Therefore, when x takes the maximum value of 20, the cost of greening has a minimum value of 360000-4000 20 280000 yuan.

==。This thing was added when the equation was first solved. Anything that is supplemented, how can it be useless. For example, the ternary equation that you added in the first semester.

In fact, solving inequalities is similar to solving one(two) one-dimensional equations, and I guess you learned the function so you think you should remedy it.

For example, 10+x 13, first follow the steps of solving the equation, if it is "=", then x=3, but here is a greater than sign, so change the equal sign to a greater than sign (because the symbol in front of x here is ) Another example, 10-x 8, you also follow the steps of solving the equation first, and get -x -2, but what, because x here is preceded by a - sign, so change the less than sign to a greater than sign.

In the case of inequality groups, since you have learned the functions, then you should know the plane Cartesian coordinate system, so you can represent those two equations on the number line, and the common part is the solution of the inequality group. At! ~

a(x-1) is greater than x+1-2 and simplified to x(a-1) is greater than (1-a).

If the solution is that x is less than -1, a-1 must be less than 0

So get a less than 1

Solution: a(x-1) x+1-2a; ax-a＞x+1-2a；ax-x＞1-2a+a；(a-1)x＞1-a；To make the solution set of inequalities x -1, it means that both sides of the inequality are divided by a negative number at the same time, so there is: a-1 0, a 1

First, turn on the shift simplification to get (a-1)x>1-a

Because the solution set of x is less than -1, a-1 must be less than 0 in order to make the two sides inverse after dividing (a-1) by the unequal sign, so it should be a-1<0, so a<1

To make any x 2+|2x-6|a, just find x 2+|2x-6|The minimum value that can be achieved, such that a< = this minimum value, guarantees that the above equation is true for a real number x (because, no matter what value x is x 2+|2x-6|will not be less than its minimum value, and thus must be greater than or equal to a).

Obviously, the key is to remove the absolute value of the judgment, to remove the absolute value of the key is the positive and negative problem of 2x-6, divided into two cases.

When 2x-6 0 (i.e. x 3), the original absolute value sign becomes x 2+2x-6=(x+1) 2-7, and apparently the minimum value reaches (3+1) 2-7=9 at x=3

And when 2x-6<0 (i.e., x<3), the original absolute value sign becomes x 2-2x+6=(x-1) 2+5, and obviously the minimum value reaches 5 at x=1.

Combining the two types of situations, you have to get everything x, x 2+|2x-6|The minimum value of is 5, so as long as a<=5 is taken, that is, the requirements are satisfied, so the maximum value of a is 5

I'm counting if positive a1 -2 a2 3 if negative has no real root, so I think, are you sure the question is okay?? It's best to ask the teacher again.

Find x 2+|2x-6|Minimum, discussed when x>=3, the left x 2+2x-6 9

When x<3, x 2-2x+6>5

So a maximum is 5

Your thoughts are on x 2+|2x-6|It's Evergrande at zero, so round off A-7.

Correct answer A 5

When x=0, you don't think about it.

Counter-evidence method false spike dan ruler set a+b>2

a +b >a guess high +(2-a) 6a -12a+8 = 6(a-1) +2 2

Contradicts a +b =2.

That is, if the inverse of the original proposition is not true, then the delay in the original proposition is not true to obtain a+b 2

The answer should be t<0 or t>6

Sub-situation: In the first case, t<0 is directly established.

Second, when t>0, first multiply both sides by 4 at the same time to get 12 2t<1, and after about minutes, we get 6 t<1

Then multiply both sides by t at the same time to get t>6

Obviously, the team has won 10 games so far.

Assuming a total of x games need to be played, then:

10/x>=30%

x<=100/3

x max is 33

That is, the maximum number of games played in the whole season is 33

And now it has been 20 races

There are 13 more.

The team has won all 13 of these games at most.

Plus the previous 10 games.

You can win up to 23 games.

Won 10 games 10 equals 30% 10 divided by 30% equals 30 games So the team has a maximum of 30 games.

x/10≤30% x≤30