If the solution set of inequalities 2x a x 3 2x 4 includes 2, find the range of values of a

The original formula is equivalent to.

2(x-a/2)|+x+3|≥2x+4

When -a 2 3, i.e., a -6, the original system of equations is solved:

When x -3, the solution set does not include (-2, +

When x -3, (x-a 2) x +3 0

The original formula can be reduced to (2x-a)+(x+3) 2x+4, i.e., 3x-a+3 2x+4, x a-1, because a-1 -3, so this segment is x-3, including (-2, +

Therefore, when a -6, the requirements are met.

When -a 2 3, i.e., a -6, the original system of equations is solved:

When x -3, the solution set does not include (-2, +

When -3 x a 2, x-a 2 0, x+3 0 can be converted to -(2x-a)+(x+3) 2x+4;x (a-1) 3, to compare the size of (a-1) 3 and a 2.

When (a-1) 3 a 2 i.e. -6 a -2, this paragraph is interpreted as ......When (a-1) 3 a 2 is a -2, this paragraph is interpreted as ......When x a 2, the solution is x a-1, and the size of a-1 and a 2 should be compared.

When a-1 a2, i.e., a2, the solution set is

When a-1 a2, i.e., a2, the solution set is

When -6 a -2, the solution set includes the total solution set includes, including (-2, + This paragraph holds.

When -2 a 2, the total solution set of the simultaneous solution includes, excludes (-2, + This paragraph is not true.

When a 2, the total solution set of simultaneous solutions includes, excludes (-2, + This paragraph is not true.

The value range of a is (- 2).

Solution: 2x-a|+|x+3|The solution set of 2x+4 includes (-2, + 2x-a|+x+3≥2x+4

2x-a| ≥x+1

2x-a∣）²x+1）²

4x²+a²-4ax ≥ x²+1+2x

3x²+-4a+2)x+a²≥0

Discriminant = (4a+2) -12a

16a²+16a+4-12a²

4a²+16a+4

4(a²+4a+1)

4[(a+2)²-3]

4 (a+2)²-12

If you make 3x +-4a+2)x+a 0, then 0 is 4 (a+2) -12 0

a+2)²≤3

3 ≤ a+2 ≤√3

2 - 3 ≤ a ≤√3-2

Summary. Hello, can I send it over?

1.The group of inequalities (x-2<3ax+4) is known

Hello, can I send it over?

Can you hurry up. Hurry up, hurry

In a hurry. Are you true or not, I already knew that I wouldn't waste 2 dollars.

a-2<0, a<2.So the range of a is less than 2, if it helps you, please give a good reputation. It's not easy to answer the question, I need your support, if you don't understand something, please ask a question in the new page.

If the solution set of the inequality (a+2) x>2a-4 for x is x<3, then the range of values for a is .

2a-4=3(a+2)

Get a=-10

The unequal sign symbol is changed to indicate a-3<0.

There are x> (2a-6) (a-3).

x>2。

a-3<0, solution a<3.

Solution: Only the interval [1,2] satisfies the equation of not sock touching |x+a|+|x-2|≤|x-4|Can.

At this time, x-2 0, x-4 0, so the original inequality is:

x+a|+2-x≤4-x

x+a|≤2

2 x+a 2 pairs of any x to accompany the elimination [1,2] is true, Lu is good to know.

Therefore, the value range of a is [-3,0].

Part Lun x=1 or x=2, the result is that the ear is raised early x-2 0, x-4 0, so.

x+a|+|x-2|≤|x-4|That is.

x+a|+|x-2|The opposite of |x-4|(the negative number and the absolute pair of 0 are his inverse).

i.e. |x+a|+2-x≤4-x

Simplification|x+a|≤2

Woo! I can only do this step, and I forgot about it

x+5-1 x+2 x+2

Isn't that right? Please use .

I don't know if you're talking about (ax+2) 2 or ax 2+2, for example 2 is 1 2, 7-2 is 7 3-2, and 7-2 is (7-2) 3

Hehehe, kid, didn't study hard, first simplify this inequality x 2+2 5-41 2 part of it, then find the intersection 1, (1-a) 2>0 to get a (1-a)> 1 2 a>-11 The second is (1-a) 2<0, get a>1

Solution: There are three scenarios to ensure the integrity of the calculation process.

1) A=2, which obviously contradicts the title.

2) A>2, a-2>0, so there is x<(2a-4) (a-2), i.e., x<2, which contradicts the title.

3) When a<2, a-2 < 0, so 2-a>0, both sides multiply -1 at the same time, there is (2-a)x>4-2a

So x>(4-2a) 2-a, i.e. x>2. Therefore, a<2 are attached.