Elementary 1 math problem, about the absolute value of the master advanced

-8|is the distance between the point representing -8 and the origin 0 on the number line, this problem is a question of examining the absolute value, in general, the standard mass is regarded as 0, and if the mass of the ball is checked to be 0, it is the standard mass 0;

If the quality of the ball is checked, it is a positive number;

If the quality of the ball is < 0, it will be negative.

That is, the closer the absolute value of the inspection quality of the ball is to 0, the better the quality.

Ball 1: +15 The absolute value is 15

Ball 2: -10 The absolute value is 10

Ball 3: +30 The absolute value is 30

Ball 4: -20 The absolute value is 20

Ball 5: -40 The absolute value is 40

From this it can be seen that the quality of Ball 2 is the best!

x|=4, depending on the definition of the absolute value, x=+4 or -4y|=3, according to the definition of absolute value, y = +3 or -3 and x>y When x = 4, y can be +3 or -3

When x=-4, y has no such value.

So x=4 y=+3 or x=4 y=-3 come on!!

The absolute value is the distance of the value from the zero point.

1 represents a point of 0.

2. Ball 2 has the best quality, and the specified mass is the point representing 0 on the number line, and the distance between the actual mass x and the point representing 0 is |x|, then |x|The smaller it is, the smaller the difference from the stated mass, because ball 2 has the smallest difference from the stated mass, so ball 2 is the best.

3，|x|=4|y|=3 so x = 4 or -4, y = 3 or -3 again because x > y

When x=-4.

x>y can't be true.

contradicts known conditions.

When x=-4.

y=3 or -3

All meet x>y

Sum up. x = 4, y = 3 or -3

-8|is the distance between the point and the zero point of -8 on the number line.

Known |x|=4,|y|=3, and x>y Please explore the value of x,y.

x=4 y=3

1.Origin. 2.Ball 2 is good because it is only 10 short of the standard amount

The quality of 2 with 0 balls is the best, because the sub-10 pair of plants is the smallest, so the quality difference between him and the standard is the least, so the quality is the best.

Because x》y, x must be positive, so x=4y=3 or =minus 3

Question 1 0 Question 2 Ball 2*** The smaller the difference between the mass of the ball and the specified mass, the better, that is, the smaller the absolute value of the given data, the better.

Question 3 x=4,y=3 x=4,y=-3

Because the absolute value of a number is.

Non-negative numbers, simply means that the absolute value of a number is at least equal to 0, so x-6|+|y+2/3|=0, only 0+0 0, so x-6|0, x=6, |y+2/3|=0 to get y -2 3,—y + x-2 3 6

The original formula can be converted into |a+b|+|a-b|=2b, by the nature of the absolute value there is b>0, so that the equation on the left is 2b, then the first.

a+b|+|a-b|It must be reduced to (a+b)+(b-a), that is, for option a, take a counterexample b=2, a=-3, obviously option a is not valid.

a,c move 2b,note,b>0;

Equal to 2b, description, a+b>0, b-a>0

So, it can be inferred from this that b>0, b>a

Options are available.

I've forgotten all the knowledge of junior high school You use the elimination method When it is, it doesn't hold, when it is c, a-b is less than 0, so 丨a+b丨+丨a-b丨=a+b-(a-b)=a+b-a+b=2b, 2b-2b=0

The answer is: c

If you ask why, it's a case-by-case discussion, stripped of the absolute value notation. It was very cumbersome to do this at the time, such as:

In case 1, when a is greater than 0, b is greater than 0, and a is greater than b, the original formula becomes: a+b+a-b-2b=0, then a=b, so a is greater than 0, b is greater than 0, and a=b is satisfied.

In case 2, when a is greater than 0, b is greater than 0, and a is less than b, the original formula becomes: a+b+b-a-2b=0, then a,b are arbitrary real numbers, but the prerequisites must be satisfied.

Case 3, etc.

Scenario 4, etc...

In fact, you only need to attach the actual numbers of A and B to this kind of problem, and you can quickly experiment with it. For example, if a=1 and b=2, then 1+2+2-1-2 2=0

You can wait until b is greater than a and is greater than 0

Special value method, substitute a few numbers, and it's OK to try it!!

|-5|=5

It means that the number "-5" is 5 (number line) units away from the origin [i.e., the "0" point on the number line).

The geometric meaning of the number "-5" is usually described as the length of the line segment from the point to the origin of the number "-5". [for "5" units].

i.e. when calculating the absolute value, only the magnitude of the "number" value is emphasized, and its directional property is discarded. [The "value" of the absolute value must be non-negative

is equal to 5, which is called the absolute value of minus 5.

The absolute value of a negative number is its opposite, which is a positive number, and the absolute value of a positive number is that it is itself a positive number.

Then the absolute value of -5 is 5, i.e., |-5|=5.。。

If you don't understand, please feel free to ask.

｜－5｜=5

The absolute value represents the distance from a point on the number line to the origin (i.e. to point 0), the distance must be a positive number, and the distance from 5 to the origin is 5, so its absolute value is 5

by 丨a丨=4, 丨b丨=5

Get a = 4 or -5

And because of B a

So a=4, b=-5

Because 丨a丨=4, then a=4, -4, because 丨b丨=5, so b=5, -5

b a, so b = -5, a = 4 or -4

There are two possibilities, a=4, b=-5 or a=-4, b=-5 analysis:|a|=4 then a=4, or a=-4

b|=5 then b=5, or b=-5

Elect b from these

Solution: (1) a, b is a positive number, and c is a negative number.

a|-|b|+|c|

a-b-c2) when m is positive, 5|m|=-5m

5m|=5m

5|m|＜|5m|

When m is negative, 5|m|=5m

5m|=-5m

5|m|＜|5m|

1.The original formula = a-b-c just needs to determine the positive and negative of these numbers.

2.The former is less than or equal to the latter because the absolute value is always greater than or equal to zero.

Original = a-b-c

2. Greater than or equal to.

Because -5 m is a negative number or zero -5m is a positive number or zero.

||Set |According to known conditions, |a+b|The value of can only be: 0,1,2|b+c|,|c+d|AND |d+a|The value of can only be: 0,1,2 assumes |a+b|=2, then |b+c|,|c+d|AND |d+a|The value of could only be:

0 So|a+b|+|b+c|+|c+d|+|d+a|=2 can be reduced to:

b+c)+(c+d)+(d+a)=0

a+b)+2(c+d)=0

a+b=0 This is the same as |a+b|=2 contradictions. So |a+b|=2 does not hold the hypothesis|a+b|=0, or 1, after trying, it is found that |a+b|+|b+c|+|c+d|+|d+a|=2 is true, so :|a+b|= 0 or 1

Hope it helps.

|, because |a+b|=a+b, so both a and b are positive, if they are both negative numbers or one positive and one negative their absolute value cannot be their sum, you can algebraically verify it, so a=3, b=1. And because |a+c|=-|a+c|, a=3, a is positive, and their absolute value is negative, so c is negative, c=-5. Then a-b+c=3-1+(-5)=-3

The distance between the point corresponding to a number on the number line and the origin (at zero) is called the absolute value of the number. Absolute values can only be non-negative. Function Definition: |a|=a（a>0） |a|=-a（a<0） |a|=0 (a=0) meaning.

The absolute value of a positive number is itself, and the absolute value of a negative number is its opposite

I really don't know how you want to ask, so I can only say that, but I think the book has made it clear enough, and it's the same as this.

|Solution: By |a+b|=a+b knows that the absolute value of a+b is itself, so a+b is a non-negative number, the same reason |a+c|=-(a+c)

Knowing that a+c is a non-positive number, it is known that :|a|=3

b|=1c|=5, knowing that a+b is a positive number, and a c is a negative number.

Causes: |a|=3>|b|1, and a+b is a positive number, so a 3 (because a 3, a+b cannot be a positive number) b=1

Causes: |a|=3<|c|5, and a+c is a negative number, so c -5 (because c 5, a+c cannot be negative).

Therefore a=3b=1c=-5

a-b+c=3-1-5=-3

|a+b|=a+b, so both a and b are positive, so a=3, b=1. And because |a+c|=-|a+c|, a=3, a is positive, and their absolute value is negative, so c is negative, c=-5. Then a-b+c=3-1+(-5)=-3