Absolute value questions of the first year of junior high school, calculation questions and answers

2.The absolute value symbol indicates that the values are greater than or equal to 0.

4.Known: |x-2|+|4-y|=0, the value that is calculated.

x-2=04-y=0

x=2y=4

5.Use the number axis to find |x-2|+|x-3|+1 minimum.

x = 2 and 1 2.

Minimum = 1 2 + 1 2 + 1 = 2 looks.

x－2｜＋｜4－y｜＝0，｜x－2｜=0 ，｜4－y｜＝0，x=2 y=4

x 2 x 3 1 When x is greater than or equal to 3 x 2 x 3 1=x-2+x-3+1=2x-4 is greater than or equal to 3*2-4=2

When x is greater than 2 and less than 3 is x 2 x 3 1=x-2+3-x+1=2

When x is less than or equal to 2 is x 2 x 3 1=2-x+3-x+1=6-2x Bring x less than or equal to 2 into it Greater than or equal to 2 x 2 x 3 The minimum value of 1 is 2

1. Because the absolute value of any number is greater than or equal to zero, the sum of the two absolute values is also greater than or equal to zero, and only the two numbers are zero, and the addition is zero, so x-2=0; 4-y=0;So x=2; y=4

2.When 2When < x<3, the original formula is equal to 2-x+x-3+1=0< p>

Since a+b+c=0, b+c is equal to negative a, and then a+c= -b let a be greater than 0b and c are both negative.

From this it can be deduced that a, greater than the absolute value b is greater than the absolute value, c

I looked for it from "New Thinking", you can look at it.

1；|x+2|+|y-3|=0 to find the value of x+y.

2：|x+7-12|x|-3|Find the value of x.

3、|x+2|+|y-3|=0 to find the value of x+y 2:|x+7-12|x|-3|Find the value of x.

Here are a few big questions.

According to the conditions of the question Example 1 The result of simplification is ( ) a) (b) (c) (d) Idea analysis From the knowable The first layer of absolute value symbols can be removed, and the second absolute value symbols can be combined and sorted out and then solved by the same method (b) Inductive comments As long as you know whether the algebraic formula of the absolute value will be exactly negative or zero is exactly negative or zero is exactly negative or zero is exactly negative or zero, you can smoothly remove the absolute value sign according to the meaning of the absolute value, which is the general idea to solve this kind of problem With the help of the teaching axis Example 2 Real number a, b, c on the number axis as shown in the figure, then the value of the algebraic formula is equal to ( ) a) (b) (c) (d) thought analysis It is easy to see from the number line, which clears the way for removing the absolute value sign Solution Original formula should be selected (c) Inductive comment This type of question type is to mark the known conditions on the number line, and with the help of the information provided by the number line, people must observe: 1 The left side of the origin is a negative number, and the right side is a positive number

Solution: (1) (+4) + (3) + (10) + (9) + (6) + (12) + (10), (3) + (9) + (6) + (4) + (12) + (10) + (10), (18) + (16) + 0, -2 (cm leakage), so the last position of the snail is on the west side of point 0, and the distance from point 0 is 2 cm;

3）|+4|+|3|+|10|+|9|+|6|+|12|+|10|, 4 + 3 + 10 + 9 + 6 + 12 + 10,54 (cm), so the snail gets a total of 54 sesame seeds;

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Because the absolute value of a number is non-negative, it 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 sum of absolute values can only be 0 x=6 y=-2 3

y+x-2/3=2/3+6-2/3=6

Since the absolute value is greater than or equal to 0, so x-6 = 0 and y+2 3 = 0, then the result is 6

The absolute value of the difference between a and b represents the distance between a and b on the number line;

To make x—1 + x—2 + x-3 +....x—617 to obtain the minimum value;

then x is in the middle of 1--617, and x=(1+617) 2=309 then x—1 + x—2 + x-3 +....x—617↓=↓309—1↓+↓309—2↓+↓309-3↓+…309—617↓

I hope it can help you, if you don't understand, please hi me, I wish you progress in your studies!

When x=309, the above equation is the smallest, and the minimum value is 2(1+2+3+. 308)

（1）x≥2

x-2|-a|=3 ⇒ x-2-a|=3 ⇒ x-(a+2)|=3

1-1）x≥a+2

x-(a+2)|=3 x-(a+2)=3 x=a+5 (this solution holds at a -3).

1-2) x3).

2-2）x<2-a

x-(2-a)|=3 x-(2-a)=-3 x=-1-a (this solution holds at a>-3).

When a=-3, the equation has 3 solutions: a+5=2

When -33, the equation has 4 solutions: a+5, -1-a, a -a

||x-2|-a|=3

Solution: Simplify the equation first.

x-2| -a=±3

x-2|=a±3

When a+3<0 and a-3<0 there is no solution to the equation, i.e., a -3, when a+3>0 and a-3<0 there are two solutions to the equation i.e., -3 a 3, when a+3>0 and a-3>0 equation has four solutions, i.e., a 3, and when a=-3 there is one solution.

When a=3 there are three solutions.

So when a=3 there are three solutions; A 3 has four solutions.

If you don't understand, please ask, and I wish you progress in your studies! o(∩_o

Solution:When x>=2, |x-2-a|=3 x=5+a>=2 has a>=-3 or x=a-1>=2 has a>=3

When x>=2, |2-x-a|=3 x=-1-a<2 has a>-3; Or x=5-a<2 has a>3

1.When a=-3, there is only one solution x=5+a=22When -33, there are four solutions x=5+a x=a-1 x=-1-a x=5-a

||x-2|-a|=3，|x-2|-a=3 or -3,|x-2|= a+3 or a-3

A+3<0 and a-3<0 have no solution to the equation i.e., a -3a+3>0 , and a-3<0 has two solutions to the equation i.e., -3 a 3a+3>0 and a-3>0 equation has four solutions, i.e. a 3a=-3 has a solution.

There are three solutions when a=3.

There are three solutions when a=3. A 3 has four solutions.

∵｜｜x-2｜-a｜＝3

x-2｜-a＝±3

i.e.: | x-2 |=3+a or | x-2 | = -3+a ②

1. When 3+a>0 and -3+a 0, the equation has two solutions, and the equation has only one solution, at this time, a=3, the solution of the equation is 8, 4, 2;

2. When 3+a>0 and -3+a>0, both equations have two solutions (i.e., the original equation has four solutions), and a>3,

|x-2|= a+3 or a-3

x = 2 + (a + 3) or 2 - (a + 3) or 2 + (a - 3) or 2 - (a - 3) i.e. x = a + 5 or -a + 5 or a - 1 or -a - 1

Obviously, a+5 is not equal to a-1, -a+5 is not equal to -a-1, then when there are only two solutions, a+5=-a+5 and a-1=-a-1, that is, a=0, and only three solutions, a+5=-a-1 or -a+5=a-1, i.e., a=3 or -3, and only four solutions, a is not equal to 0 and plus or minus 3

There are two solutions when a=0, three solutions when a= 3, and four solutions when a≠0 and a≠ 3.........

Square the two sides, and then classify the absolute value of x-2, and finally get a quadratic equation such that b2-4ac=0 has three solutions. b2-4ac is greater than 0 and there are four solutions.

||x-2|-a|= |x-2|=a+3 or |x-2|=a-3;When a+3 0 and a-3=0, x has three solutions, a+3>0, a-3>0 and a-30 x has four solutions.

1) Greater than or equal to.

When a,b>0, the left side is equal to the right side.

When a, b<0, the left is still equal to the right.

When one of a and b is greater than 0 and less than 0, the left side is greater than the right side of 2) less than or equal to 0

The one on the right can be seen as the distance from x to 2010 (on the number line) and the | on the leftx|It can be seen as the distance from x to 0, plus 2010 is obviously x to be less than 0 to see so many words, give it to me.

1) Greater than or equal to.

2 x less than or equal to 0 and greater than or equal to minus 2010

1 is greater than or equal to 0

2 Less than or equal to 0