The equation about X x 1 2x 1 a has a solution, and the range of values of the real number a is foun

No, because the front and back do not represent the distance from the same point to the other two points.

It is better to discuss it in zero-point segments.

x+1=0, the solution is x=-1

2x-1=0, the solution is x=1 2

x -1, |x+1|+|2x-1|=-x-1+1-2x=-3x 31 x 1 2,|x+1|+|2x-1|=x+1+1-2x=2-x, then 3 2 2-x 3

x 1 2, |x+1|+|2x-1|=x+1+2x-1=3x≥3/2

So |x+1|+|2x-1|≥3/2

So |x+1|+|2x-1|=a has a solution, then a 3 2

x-1|+|x+1|The sum of the distance from point x to points -1 and 1 is the shortest line segment between the two points, so |x-1|+|x+1|≥1+1=2

So a 2, the equation for x is |x-1|+|x+1|=a has a solution.

Give a little process, and say the details yourself....

First, make a graph of the absolute value function (plot the wave of the total amount...).Then to make the function a have a solution means that the line y a has an intersection with the function image. Just look at the range where A intersects with the image....

In fact, this is a question that is the most worthwhile...Just find the maximum and minimum values of the function, and the closed interval is the range of a....

Zero-point segmentation method, you can draw an image, it's not very difficult!

1. From the discriminant formula 0, 4+4a 0,a -1 is obtained.

2. It is obtained by x 2x a=0, a=x 2x=x (x-2).

0 x 2, get a 0.

x 2, get a 0.

When x=2, a=0

Combined: 1 A 0 or A 0

i.e. a -1.

Solution: x squared - 2x-a=0

x squared - 2x + 1 = a + 1

x-1) squared = a+1

x-1=√(a+1)

or x-1=- a+1).

i.e. x=1+ (a+1).

or x-1=- a+1).

x>01-√(a+1)>0

a+1)<1

and a+1>0, i.e. a>-1

1>a+1>0

0>a>-1

4+4a>=0, which is the discriminant of the root.

a>0, so -1<=a<0

Recipe: (x-1) 2 - 1+a) = 0

Solution: a = 1+(x-1) 2 1, because -1 adds a number with a minimum of zero.

Answer: f(x)=x 2+x-a, the parabola opening is upward, and the axis of return is good early x=-1 2

It is a monotonically increasing function on the area (socks 0,1).

In (0,1) there is a zero point, then there is:

f(0)*f(1)

The equation about x x 2x a=0 has a solution at 2 x 2 and <=a=x 2-2x=(x-1) 2-1(-2

There are two solutions to this problem.

One is to find the root of the equation, place the two roots between (-2,2), and then merge the conclusions.

The second is to use the zero-point theory of functions to solve the problem by combining numbers and shapes.

First, by the discriminant formula 0, obtain: a -1.

Second, the result is obtained by: x 2x=a, which is divided into three cases.

x 2, get a 0.

2, x=0, get a=0;

3. -2 x 0 gets a 0.

Combine the above. 1 and 2 gets: -1 a 0

Illustrates that there is one or two roots at (-2,2), separating a=x 2-2x, so that h(x)=x 2-2x, the opening is upward, the axis of symmetry x=1, h(x)min=h(1)=-1, h(x)max=h(-2)=8, the equivalent y=a line has an intersection with the function h(x), then -1 a 8

For example, we can see that x 2-2x-a=0 has a solid root, and the real root x (-2,2), then there is:

-2）^2-4×（-a）

4+4a≥0，4a≥-4，a≥-1；

The function f(x)=x 2-2x-a opens upward, first decreases and then increases, f(-2)>0, f(2)>0

4+4-a>0,4-4-a>0,a<8 and a<0,in summary,a [-1,0].

Equation|x|=1-2ax has and only picks up positive solutions.

x＞0x=1-2ax

x=1/(1+2a)＞0

a Imperial match -1 town refers to 2

|x-1|+|x+1|The sum of the distance from point x to points -1 and 1 is the shortest line segment between the two points, so |x-1|+|x+1|≥1+1=2

So a 2, the equation for x is |x-1|+|x+1|=a has a solution.

Solution: Because of the equation about x|x+1|+|x-1|=a has a solution, so when x -1, (x+1)-(x-1)=a,2x=a,x=-a 2,a 2,a 2 -1,a 2.

When -1 x 1, x+1-(x-1)=a, a=2, when x 1, x+1+x-1=a, 2x=a, x=a 2, a 2 1, a 2

In a geometric sense.

x-t|Represents the distance between x and t on the number line.

x-1|+|x-（-1）|Refers to the distance from x to 1 and the distance from x to -1 and =a to draw the image.

x re-1 left or 1 right, |x-1|+|x-（-1）|>2x re-1 to 1 only, |x-1|+|x-（-1）|=2|x-1|+|x-（-1）|>=2

a>=2

When x 0

The equation can be written as: 2 x-1-2 x-1=a+1 solution: a=-3

When x<0, the equation can be written as: -2 x+1-2 x-1=a+1 to solve: a=-[2 (x+1)-1] 1

Let 2 x=m;|m-1| -m+1| =a+1;(m is definitely greater than 0).

1) If 0=1, then m-1-m-1=a+1; a=-3;

In summary, a=-3u-1

Discuss when x>=0, a=-3, when x<0:-2 x-2 x=a+1 because 0<2 x<1,0<2*2 x<2

0<-(a+1)<2.-3 so. -3=

[Note: The title may be: Knowing that the equation has a real number solution, find the range of the value of a.]

Solution: t=2 xKnowable, t 0

This problem can be reduced to finding the function f(t)=|t-1|-|t+1|, (t 0). When 0 t 1, it is easy to know that f(t)=(1-t)-(1+t)=-2tAt this point, it is easy to know that -2 -2t 0

That is, the function range is [-2,0)When t 1, it is easy to know that f(t) = (t-1)-(t+1) = -2In summary, the range of the function f(t) is [-2,0).

From the question, there should be -2 a+1 0Solution: -3 a -1

The value range of a is [-3, -1).

Solution: t=2 xKnowable, t 0

This problem can be reduced to finding the function f(t)=|t-1|-|t+1|, (t 0). When 0 t 1, it is easy to know that f(t)=(1-t)-(1+t)=-2tAt this point, it is easy to know that -2 -2t 0

That is, the function range is [-2,0)When t 1, it is easy to know that f(t) = (t-1)-(t+1) = -2In summary, the range of the function f(t) is [-2,0).

From the question, there should be -2 a+1 0Solution: -3 a -1

The value range of a is [-3, -1).