High School Math Two inequality groups.

1.It is known from the meaning of the title.

2x-1>3x+2 or 2x-1<-3x+2

Then solve x<-3 or x<-1 5 and get it. x<-1/52.Or can it be known from the title.

3x-2<2x-1<3x+2

Then solve the intersection of x>-1 5 and x>-3. x>-1 5 or, and. It's a keyword.

Haven't you learned about union and intersection? If you have learned.

The first question is, the first is on a case-by-case basis, so the two ranges of the solution do not need to be established at the same time, so it is enough to combine all the solutions.

The second question, -3x-2<2x-1<3x+2, solve two ranges but must be established at the same time, otherwise how can it be greater than -3x-2, and it can be less than 3x+2, so the two solution sets must be intersected, that is, to find the common part, so as to ensure that greater than and less than are established at the same time.

You may not understand it after reading it, you can send me a short message.

Someone actually copied me.

You can go to the absolute value first, and then solve it.

1.2x-1 > 3x+2 or 2x-1<-3x-2, i.e., x<-3 or x<-1 5, so find the union, from which we get x<-1 5

2.-3x-2<2x-1<3x+2, i.e., -1 5-3, so find the intersection, from which we get x>-1 5

2x-1>3x+2 or 2x-1<-3x+2

Then solve x<-3 or x<-1 5 and get it. x<-1/52.Or can it be known from the title.

3x-2<2x-1<3x+2

Then solve the intersection of x>-1 5 and x>-3. x>-1/5

You're talking about inequality groups.

When you say "two lines on the number line coincide", do you mean finding "x 3" and "x <3" Jinghuai?

When "the two lines on the number line coincide", whether the inequality has a solution depends on whether the two lines and the two intersection points of the number axis are "hollow" or "solid", when the two intersection points are "solid", then the solution of the inequality is the "solid point"; If at least one of the two intersections is "hollow", then there is no solution.

For example, the solution of the first inequality is x 3, the solution of the second inequality is x<3, the first inequality is "solid", and the second is "hollow".

Of course, if one is x>3 and the other is x<3, then the inequality group must be unsolvable, and that should be the case you are considering.

For example, if you find the solution of the inequality is x 3, and the solution of the second inequality is x 3, these two are "solid", and their common part is x=3, so the solution of the inequality is x=3. So at this point the group of inequalities has a solution.

Yes, a single real number can also be a solution to a group of inequalities.

In a word, you conclude that you ignore the "equal sign" in the inequality group.

x/2+1/2x≥-1①

x/2+1/2x≤1②

Equals: x +1 -2x

x²+1≤2x

Transposition. x²+2x+1≥0③

x²-2x+1≤0④

Solution, yes. x+1)²≥0

x -1 solution, get.

x-1)²≤0

x 1 The set of solutions for the original inequality equation is .

1≤x≤1

Multiply the two known inequalities by 2 to become the equivalent two inequalities x+1 x

x+1/x ≤-2

Function f(x) = x+1 x

It is a function with a sign whose value range is [2, positive infinity), (negative infinity, -2] and can also do the same:

When x>0, f(x)=x+1 x 2

fundamental inequalities), a minimum of 2 is reached when x=1

When x<0, f(x)=x+1 x=-(-x+1 (-x)) 2 (basic inequality), and when x=-1 reaches a maximum of -2 Therefore, the solution set of the inequality is x=1, or x=-1, and only the equal sign holds.

Solution: 3x 2+x-2 =0 ·· 1）

4x^2-15x+9＞0···2）

From (1) to get (3x-2) (x+1) 0, that is, there is x 2 3 or x -1 from (2) to get (4x-3) (x-3) 0, that is, there is x 3 or x 3 4 combined, we have x or x -1

I didn't understand the question, multiplying by 2 or the letter 2 can be represented by y.

3x+3>1

x>-2/3

x-4≤8-2x

3x 12x 4 so -2 3 so the smallest integer solution is x=0

1. xy-x-y=1

x-1)(y-1)=2≤[（x-1）+（y-1）]^2/4(x+y-2)^2≥8

x+y 2+2 2 or x+y 2-2 2

There should be a qualifier for this problem: for x>1,y>1, the set of solutions for x+y 2+2 2 is (- 2) (then a<0, and the solution of ax 2+bx+c=0 is x1=-2, x2=-1 2x1+x2=-b a=-5 2

x1*x2=c/a=1

So b=5a 2, c=a

ax^2-bx+c>0

Let the two roots of ax 2-bx+c=0 be x1, x2x1+x2=b a=5 2

x1*x2=1

So x1=2, x2=1 2

Because a<0, so.

The solution set for ax 2-bx+c>0 is (1 2,2)3(a 2-1) x 2-(a-1)x-1<01)a 2-1=0 and a-1=0, where a=12)a 2-1<0 and =[-(a-1)] 2-4(a 2-1)(-1)<0

a^2-1<0,-1a^2-2a+1+4a^2-4<0

5a^2-2a-3<0

5a+3)(a-1)<0

3 5 At this time, -3 5 In summary, the value range of a is -3 5

1、xy-x-y=1 ，xy-x-y+1=(x-1)(y-1)=2x-1+y-1≥2（x-1)（y-1）^1/2≥2√2x+y≥2+2√2

2. The set of solutions of ax 2+bx+c<0 is , then the sum of -2 is the root of the equation ax 2+bx+c=0.

b=,c=a

and ax 2+bx+c<0, which is less than 0

And x<-2, x> corresponding to the inequality is (x+2)(x+>0, then ax 2+bx+c<0 is divided by a when the two sides of the unequal sign are redirected, so a<0

So b=<0, c=a<0

ax^2-bx+c>0

Divide ax 2-(b a) x + (c a) < 0 on both sides

x^<0

0-1<0 conforms to ii, if a=-1 formula=>x<1 2 does not match (2) a 2-1><0, i.e. a>< 1.

For x r both f(x)<0 then a 2-1<0 <0 = >-15a 2-2a-3<0 =>-3 5

q1.Transform 1 x+9 y=1 into (x-1)(y-9)=9, and ((x-1)+(y-9)) 2=((x-1)-(y-9)) 2+4(x-1)(y-9)>=36, so x-1+y-9>=6 or x-1+y-9<=-6

x+y>=16 or x+y<=4

If no other conditions are added, there will be no minimum value for x+y.

If x,y has other conditions attached, e.g., the restriction x and y are both positive, then further investigation shows that x>1,y>9, in this case, there will be x+y>=16, and x+y will have a minimum value of 16 (in this case, x=4, y=12).

q2.As long as x 2-mx+3=0 has a root, the range of y=sqrt(x 2-mx+3) is y>=0

Therefore, the discriminant formula m 2-12>=0So, m>=2sqrt(3) or m<=-2sqrt(3).

q3.At 1<=x<=2, the function decreases monotonically. (It is not difficult to reach this conclusion as long as the intervals (-inf,0],[0,1],[1,2],[2,+inf) are discussed.) ）

q4.y=1+(a-b) (x+b) is easy to obtain when a>b>0, and the function is monotonically reduced when x>-b. At the same time, when x<-b, the function is also monotonically reduced.

q5.Four boys lined up in a row, with a total of 5 empty seats in the front and back and in between. Insert the girls into the spacers, and the different arrangements are:

4!*c(5,2)*c(2,1)*3!=24*10*2*6=2880.

q6.The coefficient of a 2*b 5*c 3 is: 10!/(2!5!3!)*2^2*3^5*(-1)^3=-.

Yes? Clan's J CM? It should be ...... m

From the question to prepare for the impulse, get early:

2（70+x）＞350

70x 7560, then, 105 x 108

Root rotten boy according to the title of the inequality group:

x+70>350

70x "Collapse Calendar with Tuan Fu 7560

Ask for it yourself.