So: for 3-2x= 1, x=2 or x= 1.
(2) when X-3 >; 0, that is, x>3, and the equation is equal to | x-3+3x | = 1, then it is | 4x-3 | = 1.
Therefore, x= 1 or 4x-3=- 1 and x= 1/2 are given up because they are all less than 3.
So x=2 or x= 1.
2. let one or two numbers be x and the other be y.
X-y = 58 x 2 and the last two digits of y 2 are the same.
X 2 = Y 2+ 1 16y+58 * 58 and X 2-Y 2 is a multiple of 100. So X 2-Y 2 = Y 2+ 1 16y+.
Because it is a multiple of 100.
So 2y+58= 100, so y=2 1.
Then x=79.
So these two digits are 2 1 79.
Interviewee: JohnFromHill- doorman level 3 1 1-2 19:24.
x & gt=3，x = 1 = & gt； There is no solution at present.
3>x & gt=0, x=- 1, and there is no solution at this time.
0 & gtx & gt=- 1.5，3-x+3x= 1= >x=- 1
x & lt- 1.5，-(3-x)-3x= 1= >x=-2
So-1 or -2.
If the last two digits are equal, the last digit of these two digits can only be 1 and 9 or 6 and 4, and it will be 2 1 and 79 after the test.
Respondent: Linden 1 122- Shushenger 1 1-2 19:28.
1.x-3=0 to determine the demarcation point on the number axis, which will be discussed below:
1)x & lt； 3: 00, | 3-x+3x | = 3+2x | =1x =-1or x=-2.
2)x & gt； At 3 o'clock, | x-3+3x | = | 4x-3 | =1x =1or x= 1/2.
3) When x = 3, it is not a solution.
On the whole, only x=- 1 or x=-2 meets the requirements.
The second problem is that it's awesome to go upstairs.
Defendant: cold-blooded-the first stage of probation 1 1-2 19:34.
1.x= 1，x=- 1，x= 1/2，x=-2
2. 1 1 and 69 or 2 1 and 79
Let one number be x+58 and the other number be x.
Because only when the tail numbers are 1 and 9 (for example,1and 19)2 and 8, 3 and 7, 4 and 6, the square tail numbers of these numbers are the same.
Because the difference between these two numbers is 58, the last number of the smaller one of these two numbers plus 8 is the last number of the other two numbers, and these two last numbers also meet such requirements. When the last numbers are 1 and 9 (for example,1and 19)2 and 8,
So the last digits of these two digits are 1 and 9, so that 1+8=9 can be satisfied.
There are 1 1 and 69,21and 79,31and 89,41and 99.
According to the calculation, only 1 1 and 69 or 2 1 and 79 meet the requirements.
Interviewee: Julang-gatekeeper II 1 1-2 19:37.
The first one is (-2)
Respondent: lrjgrzh- magic apprentice level 1 1 1-2 19:40.
1.∫| | x-3 |+3x | = 1，
∴|x-3|+3x= 1 or |x-3|+3x=- 1,
(1). If |x-3|+3x= 1,
(1). When x≥3,
X-3+3x= 1, x=2, which does not meet the premise of x≥3 and should be discarded.
When x < 3,
3-x+3x= 1，x=- 1。
(2). If |x-3|+3x=- 1,
When x≥3,
X-3+3x=- 1, x= 1/2, which does not meet the premise of x≥3, should be abandoned.
When x < 3,
3-x+3x=- 1，x=-2。
X =- 1 or -2
These two numbers are 2 1 and 79.
Let the smaller number be x and the larger number be (X+58).
(x+58)^2-x^2= 1 16x+3364，
Because the last two digits of the squares of these two numbers are the same, the last two digits of the difference of their squares are 00, that is, the last two digits of 1 16X+3364 are 00, so the last two digits of1/6x are 36, so we can see that X = 2 1.
Responder: cloudy and frosty-magician level 5 1 1-2 19:45.
forehead
Respondent: Jing 7979- Magic Apprentice Level 1 1 1-2 19:47.
1.
||x-3|+3x|= 1
|3-x+3x|= 1
3+x-3x= 1
3-2x= 1
-2x=-2
x= 1
Think about the second question.
Respondents: 16 16 Zhong-Apprentice Magician II 1 1-2 20: 12.
Read more review books
Respondent: xujianan 448- Apprentice Magician II 1 1-2 20: 14.
Not challenging.
Respondent: Gumu Xidai-Scholar II 1 1-2 20:23.
It's too difficult! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! It is beyond my knowledge.
Responder: Dream of Frost Dragon-Trainee Magician Level 2 1 1-2 20:43.
aaaaaaaaaaaaaaaaaaaa
Interviewee: Guo 1994- assistant level 3 1 1-2 20:44.
The answer is-1, which holds when it is -2.
Held on 2 1 79.
Respondent: 99 104252- probation level 1 1 1-2 20:47.
Several people upstairs answered very well, so I don't have to teach fish to swim!
Interviewee: jxs _ 8- Scholar Level 3 1 1-2 20:56
1.-2 or 1
2.79 and 2 1
Answer: Dark Thief-Magic Apprentice Level 111-221:15.
This kind of problem is the first day of the city.
Answer: moonlight in winter-trainee magician level 2 1 1-2 2 1:26.
simple
Interviewee: seaxiaohan- junior 1 1-2 2 1:43.
exam-oriented education
Interviewee: ylice- Assistant Level 2 1 1-2 2 1:52.
1, (1) When X≥3, because X-3≥0, || x-3 |+3x | = | x-3+3x | = | 4x-3 | = 4x-3 =1.
Solution: X= 1
Because X≥3 and X= 1 are contradictory, they are not attached.
(2) when x
At this time, it is divided into two situations:
① When 3+2X≥0, that is, X≥-3/2, |3+2X|=3+2X= 1.
Solution: X=- 1
Connection condition: -3/2 ≤ x
② When 3+2x
Solution: X=-2
Attachment condition: x
To sum up, the solution of equation ||x-3|+3x|= 1 is X=- 1 or X=-2.
2、
Note: If the difference between two digits is 58 and the last two digits of its square are equal, the single digits are only _9 and _ 1, _4 and _6. So these two digits can be set to 10X+9, 10A+ 1 or 65438. Then the last two digits of the square of these two digits are 180X+8 1, 20A+ 1, or 80X+ 16, 120A+36 respectively.
①( 10X+9)-( 10A+ 1)= 58
180 x+8 1 = 20A+ 1
Solution: both x and a are negative numbers, which does not meet the meaning of the question.
②( 10X+4)-( 10A+6)= 58
(80X+ 16)=( 120A+36)
Solution: x and a are decimals, which do not meet the meaning of the question.
In the end, there is no such double digit.
Respondent: Yulanwu-Juren Grade 4 1 1-2 2 1:59.
1.( 1) When x-3 is less than 0, that is, x < 3, the equation is equal to | 3-x+3x | =1:| 3-2x | =1.
Then: for 3-2x=- 1, x=2 or x= 1.
(2) when X-3 >; 0, that is, x>3, and the equation is equal to | x-3+3x | = 1, then it is | 4x-3 | = 1.
Therefore, x= 1 or 4x-3=- 1 and x= 1/2 are given up because they are all less than 3.
So x=2 or x= 1.
2. let one or two numbers be x and the other be y.
X-y = 58 x 2 and the last two digits of y 2 are the same.
X 2 = Y 2+ 1 16y+58 * 58 and X 2-Y 2 is a multiple of 100. So X 2-Y 2 = Y 2+ 1 16y+.
So 2y+58= 100, so y=2 1.
x=79
So it's 2 1 79.
Responder: Kelly PC- Magic Apprentice Level 1 1 1-2 22:25.
1.( 1) When x-3 is less than 0, that is, x < 3, the equation is equal to | 3-x+3x | =1:| 3-2x | =1.
So: for 3-2x= 1, x=2 or x= 1.
(2) when X-3 >; 0, that is, x>3, and the equation is equal to | x-3+3x | = 1, then it is | 4x-3 | = 1.
Therefore, x= 1 or 4x-3=- 1 and x= 1/2 are given up because they are all less than 3.
So x=2 or x= 1.
2. let one or two numbers be x and the other be y.
X-y = 58 x 2 and the last two digits of y 2 are the same.
X 2 = Y 2+ 1 16y+58 * 58 and X 2-Y 2 is a multiple of 100. So X 2-Y 2 = Y 2+ 1 16y+.
Because it is a multiple of 100.
So 2y+58= 100, so y=2 1.
Then x=79.
So these two digits are 2 1 79, so they are calculated.