The first one dimensional inequality group, calculation questions, 60 questions, with answers

(1) 2x-4 x+2 and x 3 solutions are 3 x 6 (2) 2x-1 1 and 4-2x 0 solutions are unsolved.

3) 3x+2 5 with 5-2 1 solution set is 1 x 2 (4) x 1 2 with 2x+3 2+x solution set is -1 x 3

5) x+3 1 and x 2 (x-1) 1 solution set is -2 x 1 (6) 2x+1 3 and x -3 solution set is 1 x -3

7) 2x+5 1 and 3x+7x 10 solution set as 1 x 2 (8) 2x-1 x+1 and x+8 4x-1 solution set as x 3

9) 1-2(x-1) 5 and 2 (3x-2) x+1 2 solution set is -1 x 3 (10) 2x 4+x and x+2 4x-1 solution set is 1 x 4

As long as the solution of the inequality group can solve the inequality, after each inequality is solved, the inequality group is the solution that they hold at the same time, so as long as the solution of each inequality intersects.

Unary inequalities can all be written.

ax>b (where a≠0).

(not equal signs"<"is the same). If a>0, you can divide a by both sides at the same time, which does not change the direction of the inequality, so there is.

x>b/a

That's the solution. If a<0, the direction of the inequality changes after dividing by a, so x44x+20>0

The first equation has, 2x<-1, solution.

x<-1/2

The second formula takes a negative sign on both sides, and the direction changes, thus.

x<-4

The third equation has 4x>-20, divided by 4, thus.

x>-5

If three inequalities are satisfied at the same time, the group of inequalities is satisfied. The intersection of the three solutions is the solution of the group of inequalities, so the final solution is.

5

Take a pick here and maybe it helps!

If the number of students is n, then: (3n+3) 5<3, n<4, take 3,2,1 test: n=3, then the books are 12, and each student in front of them is divided into 5 books, then the last person is 2 books, which is eligible;

n=2, then the books are 9 books, each student is divided into 5 books, and the last person is 4 books, which do not meet the requirements and are discarded;

n=1, which does not agree with the ruler of the topic, and is discarded;

1) 3-(a-5)>3a-4 (a<3)2) -5x+3<2x+1 (x>1 and 1/3)3)3-4[1-3(2-x)] greater than or equal to 59 (x less than or equal to -3).

4) 6 (1-3/1x) greater than or equal to 2+1/5 (10-15x) (x greater than or equal to -2).

5) 7x-13/6> 3x-8 (x>-3)6)4x-10<15x-(8x-2) (x>-4)7) x-2-2-x> x-2/3 (x>2)8) x-2/6-x-4x-3 Greater than or equal to 0 (x less than or equal to 4).

9) x-1/3 - x-1 <1/2

10)2(5-3x)>3(4x+2)

11) 1-1/2 x >2

12）7x-2(x-3)<16

13)3(2x-1)<4(x-1)

14) 2-6 (x-5) greater than or equal to 4 (3-2x) 15) 7 + 3 x < 5 + 4x

16)5-x(x+3)>2-x(x-1)

17) x-2 (1/2 of x+2) less than or equal to 1-3 (1-x) 18) 3 (x-1) + 2 (1-3x) < 5

19) 1/3x-120) 6 (1-2/3x) <2+1/5 (10-15x) Is this okay?

15x2x greater than or equal to 400x3 10x3x less than 15002x1 greater than x1 x 108 less than 4x15x 102 greater than 3 (x1) o, 5x11 less than or equal to 711, 5x2x greater than 11x. x ten 2 is less than 4 x one 1

x a 5 is greater than 1 ten 2x. 3x 102 is less than or equal to 4x, and lowercase x is equivalent to a multiplier sign.

Horizontally in a group.

Xiao Ming went to the store to buy a photo album, there are two varieties of A and B for choice, A album is 15 yuan each, 75 photos can be loaded, B album is 10 yuan each, 45 albums can be installed, Xiao Ming has 80 yuan, to make the purchased album can hold 380 **, then how many albums should he buy A and B?

Solution: (1) Set up the purchase of C type TV x sets, type A TV sets for 4x sets, and B type TV sets for (108-x-4x) sets.

1000 4x 1500 (108 5x) 2000x 147000 Solution: x 10

Because x can only be a positive integer, the solution that satisfies the condition is: x=10, so the mall purchases at least 10 C TVs.

2) From the meaning of the title:

4x≤108-x-4x ①

x≥10 ②

By x 12

The solution set for the inequality group is 10 x 12

x can only take positive integers The solution that satisfies the condition is: x

Therefore, there are 3 purchase options:

Plan 1: 40 units of type A, 58 units of type B, and 10 units of type C.

Scheme 2: 44 units of type A, 53 units of type B, and 11 units of type C;

Plan 3: 48 units of type A, 48 units of type B, and 12 units of type C.

(1) Solution: The shopping mall has purchased at least X sets of Type C TVs, with 4x sets for A and 108-5x sets for B.

4x*1000+x*2000+(108-5x)*1500≤147000

Solution x 10

A: The mall has purchased at least 10 Type C televisions.

2) The number of TV sets of type A is required not to exceed the number of TV sets of type B, then 4x 108-5x, and the solution is x 12

Combine (1) to give 10 x 12

So there are three schemes: when x = 10, A is 40 units, B is 58 units, C is 10 units when x = 11, A is 44 units, B is 53 units, C is 11 units when x = 12, A is 48 units, B is 48 units, and C is 12 units.

Solution: (1) If the mall purchases x sets of type C televisions, then there are 4x sets of type A televisions and (108-x-4x) sets of type B televisions, which is based on the title.

1000×4x+1500×（108-x-4x）+2000x≤147000

4000x+162000-7500x+2000x≤147000

1500x≤-15000

x 10 means that the mall has purchased at least 10 Type C televisions.

2) From (1), it can be seen that there are 4x sets of TV sets in type A and (108-x-4x) sets in type B TV sets.

4x≤108-x-4x

9x≤108

x 12 is in turn made up of (1), x 10

10≤x≤12

x must be a positive integer.

x = 10 or 11 or 12

There are three purchase options:

Purchased 40 units of Category A, 58 units of Category B, and 10 units of Category C.

Purchased 44 units of Category A, 53 units of Category B, and 11 units of Category C.

Purchased 48 units of Category A, 48 units of Category B, and 12 units of Category C.

Note: (1) When a column of unary inequalities (groups) solves such problems, a specific quantity must be set when setting unknowns, that is, words such as "at least" and "most" cannot be used.

2) According to the way of questioning in the question, determine whether the final solution of the unknown is required or a specific solution. For example, the first question of this question is a solution set, and the second question is a specific solution.

It took 10 hours for a steamship to return to A at a constant speed from a place A in the upper reaches of a certain river, and less than 12 hours to return to A at a constant speed from a certain river.

When it is 3 km, the hydrostatic velocity v of the round-trip ship is unchanged, what conditions does v meet?

The speed from A to B is (V+3)km h

The speed from B to A is (V-3)km h

Then let a and b be separated by s, then s (v+3)*10s (v-3)<12