It is known that for a rational number x there is x 3 x 2 z 2 y 3 13 y 1 z 6 , find

This question may seem complicated, but it is not difficult to solve it step by step.

Equations with absolute values must be discussed. x, y, z have three possibilities, x"-2,-2"x"3,x"3"3,"3"3"y"1,y>1,z<-6,-6-2

The maximum value of x+y+z in autumn is the case where x, y, and z are all the largest or smallest, and when x, y, z, and all are maximum, the equation becomes x-3+x+2+z+2+y+3=13-y+1-z-6

Simplified to x+y+z=8, so the maximum is 8, and the minimum value is also calculated in this way, hope.

Equations with absolute values must be discussed.

x, y, z have three possibilities, 1 x "-2, -2" x "3, x" 32 y" -3, -3 "y " 1, y>1 , 3 z<-6, -6-2

Find the maximum value of x+y+z, that is, when x, y, and z are all the largest or smallest numbers, and when x, y, z, and all are maximum, the equation becomes.

x-3+x+2+z+2+y+3=13-y+1-z-6, i.e. 2x+4+y+z=8-y-z

Simplified to x+y+z=4, so the maximum is 2.

When x, y, z, are all the smallest, the equation becomes 3-x-x-2-z-2-y-3=13-1+y+z+6, i.e., -4-2x-y-z=18+y+z

Simplified, -x-y-z=11, so the minimum is -11.

So the maximum is 2 and the minimum is -11.

2.It is known that the rational number x satisfies |x+3|+|x-10|=15, the rational number y is such that |y-3++1y+2|+|The value of y-5 is the least +.

Hello, glad to answer for you, 2It is known that the rational number x satisfies |x+3|+|x-10|=15, the rational number y is such that |y-3++1y+2|+|The value of the fiber sold by Y-5 is the smallest + the solution process is as follows: |x+3|+|x﹣10|The minimum value is 13, |x+3|+|x﹣10|15, x 3 1 4 or x 10+1 11, |y﹣3|+|y+2|+|y﹣5|The distance between y and 2,3,5 and the minimum when y 3 is represented on the axis of the hail and there is a minimum value of 7, x y 11 or x y 8;

Summary. 1.From the conditions given, it is possible to deduce that the rational numbers x, y, z satisfy:

x-3|+|x-6|=4，|y+1|+|y-2|+|y-3|=6，|z-1|+|z-5|=6, it is known that the rational numbers x,y,z satisfy |x-3|+|x-6|+|y+1|+|y-2|+|y-3|+|z-1|+|z-5|=11, then algebra + number +

1.From the conditions given, it is possible to deduce that the rational numbers x, y, z satisfy:x-3|+|x-6|=4，|y+1|+|y-2|+|y-3|=6，|z-1|+|z-5|=6, i.e. x=3 or x=6, y=-1, 2 or 3, z=1 or 5, the algebraic solution of x,y,z can be obtained:

x = 3 or 6, y = -1, 2 or 3, z = 1 or in addition, by the given conditions, the rational numbers x, y, z can be deduced satisfiing: |x-3|+|x-6|=4，|y+1|+|y-2|+|y-3|=6，|z-1|+|z-5|=6, i.e., x-3 = 4 or -4, y+1 = 3 or -3, y-2 = 3 or -3, z-1 = 5 or -5, the algebraic solution of x,y,z can be obtained: x=7 or -1, y=4 or -4, z=6 or -4

It is known that the rational numbers x,y,z satisfy |x-3|+|x-6|+|y+1|+|y-2|+|y-3|+|z-1|+|z-5|=11, then the maximum value of the algebraic equation xyz is , and the minimum value is .

The maximum value is -5 and the minimum value is -15

That's how I got it.

How did you get it?

Because |x-3|+|x-6|=4, x=3 or x=6;|y+1|+|y-2|+|y-3|=6, you can push out y=-1 or y=2 or y=3;|z-1|+|z-5|=6, you can push out z=1 or z=5. So, the rational numbers x,y,z satisfy: x=3 or x=6, y=-1 or y=2 or y=3, z=1 or z=5.

It is known that the rational numbers x,y,z satisfy |x-3|+|x-6|+|y+1|+|y-2|+|y-3|+|z-1|+|z-5|=11, then the maximum value of the algebraic equation xyz is , and the minimum value is .

What is the process.

As can be seen from the title, |x-3|+|x-6|+|y+1|+|y-2|+|y-3|+|z-1|+|z-5|=11, i.e., x+y+z=11. Let x=a, y=b, z=c, then a+b+c=11, and the solution is a=11-b-c, and a is substituted for |x-3|+|x-6|Gotta |11-b-c-3|+|11-b-c-6|=11, the solution is b+c=8, and b,c is substituted into |y+1|+|y-2|+|y-3|+|z-1|+|z-5|, the solution gives b=3, c=5, so x=11-3-5=3, y=3, z=5, so the maximum value of xyz is -15, and the minimum value is -5.

Summary. Analysis According to the nature of the absolute value, the |x+1|+|x-2|，|y-1|+|y-3|，|z-1|+|z+2|Solution: When x -1, y+(x+1)-(x-2)=-2x+1 3, when -1 x 2, y=x+1-(x-2)=3, when x 2, y=x+1+x-2=2x-1 3, so we can know |x+1|+|x-2|3. The same can be obtained:

y-1|+|y-3|≥2，|z-1|+|z+2|3, so (|x+1|+|x-2|）（y+1|+|y-2|）（z-3|+|z+1|2 3 3 = 18, so |x+1|+|x-2|=3，|y-1|+|y-3|=2，|z-1|+|z+2|=3, so the maximum value of -1 x 2,1 y 3,-2 z 1, x+2y+3z is: 2+2 3+3 1=11, and the minimum value is: -1+2 1+3 (-2)=-5 Comments This question mainly examines the nature of absolute values, and the value range of x, y, and z is the key to solving the problem according to the meaning of the question

The maximum value is

8.The rational numbers x,y,z satisfy (lx+2|+x-4((y-2|+[y-5(|z-2+z+3)=90, then (x-2y+3z).

8.The rational numbers x,y,z satisfy (lx+2|+x-4((y-2|+[y-5(|z-2+z+3)=90, then (x-2y+3z).

The maximum value is

8.The rational numbers x,y,z satisfy (lx+2|+x-4((y-2|+[y-5(|z-2+z+3)=90, then (x-2y+3z).

The maximum value is

8.The rational numbers x,y,z satisfy (lx+2|+x-4((y-2|+[y-5(|z-2+z+3)=90, then (x-2y+3z).

The maximum value is

8.The rational numbers x,y,z satisfy (lx+2|+x-4((y-2|+[y-5(|z-2+z+3)=90, then (x-2y+3z).

The terms of the flat section and the absolute value of the annihilation are both constant and non-negative.

x-1=02x-y=0

3x-z=0

The solution is x=1, y=2, z=3

x+y+z=1+2+3=6

The absolute value of any number must be non-negative.

If the sum of the two absolute values is equal to 0, then the two absolute values of Zhaofeng Chan are equal to 0

The dust of the family is x=3, y=-15, 3x+2y=-21

Square it and find the correspondence.

x+y+z=3,2√xy=2,2√yz=3,2√zx=6x=1 y=1/2 z=3/2

xyz=3/4

Move the item first, get |x-3|+|x+2|+|z+2|+|y+3/+|y-1|+|z+6|, =13 In a geometric sense, x-3 + x+2 is the sum of the distances from x to 3 and -2, and the length of the interval (-2, 3) is 5

z+2 + z+6 is the sum of the distances from z to -2 and -6, and the length of the (-6, -2) interval is 4

y-1 + y+3 is the sum of the distances from y to 1 and -3, and the length of the interval (-3,1) is 4

So x between (-2,3), z between (-6,-2), and y between (-3,1) satisfies the equation.

xyz takes the left end of the interval at the same time, that is, x takes -2, y takes -3, z takes -6 to get the minimum value of x+y+z -11, and the same way takes the maximum value of the right end of the interval 2

Got it?

|x-3|+|x+2|+|z+2|+|y+3|+|y-1|+|z+6|=13

By |a|+|b|>=|a-b|Knowable (absolute value inequality)|x-3|+|x+2|+|z+2|+|y+3|+|y-1|+|z+6|>=|x-3-x-2|+|z+2-z-6|+|y+3-y+1|=13

Only the equal sign can be taken, and the sufficient and necessary conditions for the equal sign are -2<=x<=3,-3<=y<=1,-6<=z<=-2

xyz can take the boundary at the same time, so as to derive the maximum value.

The equal sign to the condition is obtained by categorical discussion).