When x 1, the inequality x 1 root number x 1 m x 2 is constant, asking the maximum value of the rea

I don't know that people born in 97 have already done this kind of question, dare to ask if the world can still be saved? Hehe.

This question is transferred first.

x+1|+|x-2|-√x-1)≥m

Find the maximum value of m, that is, find |x+1|+|x-2|- x-1) minimum.

Now let's take a look|x+1|+|x-2|This part is equivalent to the sum of the distances between x and the two points -1 and 2 on the number line.

Well, when 1 x 2, |x+1|+|x-2|Constant is 3, then when x=2, - x-1) takes the smallest, |x+1|+|x-2|- x-1) The minimum value is 2

When x>2, set |x+1|+|x-2|-√x-1)=2x-1-√（x-1)=h(x)

2x-1 = x+(x-1)> x-1), so the larger the x, the greater h(x).

h(x) is the smallest h(2)=2

So the maximum value of m is 2

1≤x≤2

x+1-√(x-1)≥m-(2-x)

m≤3-√(x-1)

m max = 3

x≥2x+1-√(x-1)≥m-x+2

m≤2x-√(x-1)-1

Let (x-1)=t ,x=t 2+1 t 1m 2t 2-t+1=f(t), f(t) 2m has no maximum.

A belongs to (root number 2 2, positive infinity).

Number combination: Establish a plane Cartesian coordinate system.

To the left of the less than sign is a semicircle centered on the origin and with a radius of 1 above the x-axis.

To the right of the less than sign is a straight line with a slope of 1 and a longitudinal intercept.

The meaning of inequality is that a semicircle is established continuously below a straight line.

Through the slippery town to draw a picture, let Bi Ke know the above results.

The inequality x 2-mx+1 0 is constant for all x to (0,1 2).

i.e. m (x 2+1) x to everything x belongs to (0,1 2] constant holds.

So m is less than or equal to the minimum value of the right-end function.

And (x 2+1) x decreases monotonically on Sensan (0,1 2), so when x=1 2, the minimum value of 5 2 is taken

Thereupon m 5 2

i.e. the maximum value of m, this Pond is 5 2

i.e. mx<-(4+x 2).

m<-(x+4/x)

And there is x+4 x>=2 root number (x*4 x)=4, which is obtained when x=4 x, x=2"="That is, there is a monotonically decreasing on (1,2).

Therefore, the maximum value of x+4 x is 1+4 1=5

Therefore, the minimum value of -(x+4 x) is -5

m<-(x+4 x) is constant, then m<=-5

Finding the maximum value by using the mean inequality is a commonly used method of finding the maximum value, but when using the mean inequality to find the maximum value, we must pay attention to three conditions at the same time, that is, "one positive, two definite, three equal"."One positive" means that each item must be positive, "two fixed" means that the product of each item or the sum of each item is a fixed value, and "three equal" means that each item can be taken to an equal value. Ignoring any of these conditions will result in an incorrect solution.

According to the landlord's calculation x 2=2 (1-x) using the mean inequality of the time bend imitation x 2+2 (1-x)>=x (1-x) how to approximate it。。。

The actual method is that the original formula becomes a state of 1-x 2+2, burying friends 1-x+x-1 2, and then 1-x 2=2 1-x can be made.

A belongs to (root number 2 2, positive infinity).

Inclusion of numerical knots: Establish a planar Cartesian coordinate system.

To the left of the less than sign is a semicircle with the origin as the center and a radius of 1 above the x-axis.

To the right of the less than sign is a straight line with a slope of 1 and a longitudinal intercept.

The meaning of inequality is that a semicircle is established continuously below a straight line.

The above results can be seen by drawing. Do raids.

For constant to hold, m must not exceed the minimum value on the left.

List the roots of each absolute value on the left: 1,2,9,9,10,11, according to the symmetry, when x=9, the minimum value on the left is 8+7+0+1+2=18, so m<=18, so the maximum value of m is 18.

Obviously, when x = 9 on the left, the minimum value is obtained.

When x=9, the left = 8 + 7 + 1 + 2 = 18

When m=18, the original formula is constant.

Solution: |x+1|+√x-1)+|x-2|m1 x<2, x+1+2-x+ (x-1) m

x-1)+3≥m

The left side of the inequality is monotonically incremented, and when x=1, the minimum value is obtained. For any x on [1,2), the inequality holds, then.

m≤√(1-1)+3=3 m≤3

2x-1+ (x-1) m

The left side of the inequality increases monotonically, and when x=2, the minimum value is obtained. For any x on [2,+2), the inequality is true, then.

m 2 2-1 + (2-1) = 4 m 4 summed up, m 3 is obtained

The maximum value of m is 3

The first floor knew about the classification discussion, but the conclusion was wrong.

The solution on the second floor is fundamentally wrong, and this question needs to be discussed by category. If you don't discuss it, you can do it, you need to use the number line method to get it.

x+1|+|x-2|≥3

|x+1|+ root number under (x-1) m-|x-2|, i.e., m<=|x+1|+|x-2|+ root number (x-1), and x 1, so |x+1|+|x-2|>=3, under the root number (x-1)>=0, so |x+1|+|x-2|The minimum value under the + root number (x-1) is: 3, so the maximum value of m is: 3.

(1) When 1 x 2:

x+1|+√x-1)≥m-|x-2|

x+1+√（x-1)≥m+x-2

m (x-1)+3, when x = 2, m has a maximum: m = 4, 2) when x 2:

m does not have a maximum value.

The deformation is m |x+1|+(x-1 under the root number) +|x-2|The right formula is f(x).

Let's find the minimum value of f(x), the maximum value of m is this value, when 1 x<2, f(x)=x+1+(x-1 under the root number)+2-x=3+(x-1 under the root number) 3

When 2 x, f(x)=x+1+(x-1 under the root number)+x-2=2x-1+(x-1 under the root number) 4

So the minimum value of f(x) is 3

So the maximum value of m is 3

Discuss when 1 x 2.

m (x-1)+3 4 max. 4

When x>2. There is no maximum.

This problem can be solved in two ranges, because the premise is x 1, so x can be solved in two parts, one is 1 x 2, and the other is x 2

When x spring repentance group (1,2), the inequality x + mx 4 0 is constant.

That is, there is MX "Grilled Orange - (4+x 2)".

m<-(x+4 pre-calendar x).

And there is x+4 x>=2 root number (x*4 x)=4, which is obtained when x=4 x, x=2"="That is, there is a monotonically decreasing on (1,2).

Therefore, the maximum value of x+4 x is 1+4 1=5

Therefore, the minimum value of -(x+4 x) is -5

m<-(x+4 x) is constant, then m<=-5