The absolute value of the first year of junior high school improves the practice questions, urgent!!

1、(2a-1)^2≥0 |b+1|≥0

a=1/2 b=-1

1/a)^2+(1/b)^2=4+1=5

2. a>0 b>0 c>0 result is 3

One of them is less than 0 and the result is 1

Two of them are less than 0 and the result is -1

a<0 b<0 c<0 results in -3

3、|x-100/221|+|x+95/221|=-x+100/221+x+95/221=195/221=15/17

4. Non-negative integers whose sum is equal to 1 are only 0 and 1

ab|=0 |a+b|=1

a=0 then b=1 or -1 (b=0 then a=1 or -1)5, the value of the formula is constant, then the coefficient of x is 0

Requirements|4-5x|+|1-3x|=4-5x+(-1+3x)=-2x+34-5x>0 1-3x<0

The solution of 1 3 constant is 3+6=9

6、∵abcd/|abcd|=1

The number of negative numbers and the number of positive numbers in abcd are both even or 0abcd |abcd|=-1=(-abcd/|abcd|2001a] a+[b] b+[c] c+[d] d=4 or 0 or -4 then the result is 3, -1, -5

7. Let x=998 be brought in, and the coefficient of x can be 0 after removing the absolute value, and the minimum value 996004 can be obtained

2.-1, -3, 1 or 3

b=1;a=0,b=-1;a=1,b=0;a=-1,b=0;a=-1,b=1;a=1,b=-1

x<=, the constant is 9

6.5, 1 or -3

I wrote some upstairs during the process

b = -1 substitution.

Khan is dead, so make one and see again tomorrow.

If so, find the value of 2x+y.

Analysis and Solution: The absolute value of any rational number must be non-negative" can be obtained by using this feature.

The sum of two non-negative numbers is 0, and there is only one possibility: both non-negative numbers are 0.

x is a rational number, then the minimum value of x-2009 + x+2010 is (analysis and solution: "the absolute value of any rational number must be a non-negative number" using this feature can be used to obtain x-2009 0

x+2010｜≥0

There is only one possibility to find the minimum value: if both non-negative numbers are 0, then x-2009 =0, x+2010 =0, i.e.

The minimum value of x-2009 + x+2010 is 0+0=0, so the minimum value of x-2009 + x+2010 is 0.

Calculate -7 + 2 -4

Solution: Original formula = -7 + (-2)-4

The absolute values are all greater than or equal to 0, so the above equation requires to be equal to 0, so there is.

a-4|=0

b-2|=0

So there is a=4, b=2

The first question, because |a|=4，|b|=3, so a is plus or minus 4, b is plus or minus 3, but a > b, so a can't be -4, so a=4, b = plus or minus 3.

Question 1, because the absolute value of any number except zero is positive, only when x=0, |x|+13 has a minimum value of 13.

Question 2 2. Same as above, when x=0, 2-|x|There is a maximum value of 2.

In the third question, because a-1 +(a-1)=0, a-1 and (a-1) are opposite numbers to each other, so a-1 =-(a-1), so the range of values of a is a<1

The second volleyball (-10) is of better quality because the important is closest to the specified weight.

One. a = 4, b = 3 or a = 4, b = -3

Two. , minimum value 13, maximum value 2

Three. Range A to 1

The second one is good.

1.First of all, extract the subductive number to 3, 1+2+3+....33-3*（1+2+3+…33）=-2（1+2+3+…33）=-2*（17*33）

2.A parts a + b points b + c parts c - abc parts abc if all three are positive, then the above equation = 1 + 1 + 1-1 = 2 if only 2 are positive, then abc < 0, above equation = 1 - (-1) = 2 if only 1 is positive, then abc > 0, above equation = -1-1 = -2 If all three are negative, then above equation = -1-1-1 + 1 = -2 so above equation = plus or minus 2

2.A parts a + b points b + c parts c - abc parts abc if all three are positive, then the above equation = 1 + 1 + 1-1 = 2 if only 2 are positive, then abc < 0, above equation = 1 - (-1) = 2 if only 1 is positive, then abc > 0, above equation = -1-1 = -2 If all three are negative, then above equation = -1-1-1 + 1 = -2 so above equation = plus or minus 2

1.17 times 19 times -2=

2.abc 3 are all negative, then =-2;2 is a negative number then =-2;1 is a negative number then = 2;0 negative numbers = 2

1.First of all, extract the subductive number to 3, 1+2+3+....33-3*（1+2+3+…33）=-2（1+2+3+…33）=-2*（17*33）

Absolute value.

1. Multiple choice questions.

The absolute value of 1 is the smallest number ( ).

a does not exist b 0 c 1 d 12 when a negative number gradually becomes larger (but still remains negative) ( ) a its absolute value gradually increases.

b Its opposite number gradually increases.

c Its absolute value gradually decreases.

d The absolute value of its opposite number gradually increases.

2. Fill in the blanks.

1.If | －1|=0, then = if|1－ |=1, then =

2 The reciprocal of a number is itself, and the number is the opposite of a number is itself, and the number is

3 If the opposite of is 5, then the value of is 4 A number is 10 smaller than its absolute value, then the number is 5 If , and , then

3. Answer questions.

1 Fill-in-the-blank questions.

1) The sign is the number, and the absolute value is the number is 2) The symbol is the number, and the number with the absolute value is 3) The sign of 85 is the absolute value is

4) The absolute value of (5) is equal to (6) The number of absolute values equal to is

You just need to know that whether it is a positive or negative number, the absolute value is positive.

Absolute value Positive number does not change Negative number gets the opposite number 0 and so 0 The absolute value has | |

No!!!

I'm still looking for this question.

Directly remove the absolute value symbol, and swap the positions of the two numbers inside, that is, take the right.

l1/3-1/2l +l1/4-1/3l+ l1/5-1/4l+……l1/10-1/9l

1 2-1 10 (except for the first and last two numbers, the rest of the two connected numbers are offset to zero) = 2 5

1）|a-1|-|a-2|=1-a-(2-a)=1-a-2+a=-1The key to this problem is to figure out whether the absolute value sign is positive or negative, such as |a-1|A-1 is negative, so |a-1|The opposite number 1-a. is equal to a-1

2) Cause|a|=3，|b|=6, then a = 3, b = 6, and because a and b have different signs, a = 3, b = -6 or a = -3, b = 6

The distance between two points represented by the two numbers ab on the number line is 94)|x|-5 2=0, then |x|=5 2, so x= 5 2

Question 1 -1

Question 2 A is plus or minus 3 b is plus or minus 6 When a = 3, b = -6 When a = -3, b = 6 The distance between two points is 9 units.