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3x + 4 / (X-2) (x + 1) = A / (X-2) - B / (x + 1), where a and B are constants 4A-B／4A+B
On the right side of the equation, we obtain (AX + a-bx + 2b) / [(X-2) (x + 1)],
Therefore, the molecule ax + a-bx + 2B = 3x + 4 of the above formula;
So A-B = 3;
A+2B=4；
You'll be in the back
Put 1, 2, 3 The 100 natural numbers, 100, are randomly divided into 50 groups with two numbers in each group. Now, record any value of the two numbers in each group as a and the other as B, and substitute them into the algebraic formula 12 (| A-B | + A + b) to calculate the result. After 50 groups of numbers are substituted, 50 values can be obtained, then the maximum sum of the 50 values is___ ．
① If a ≥ B, then the absolute value sign in the algebraic formula can be directly removed, and the algebraic formula is equal to a. ② if b > A, then the absolute value sign is opposite, and the algebraic formula is equal to B. thus, if you input a pair of numbers, you can get the largest number in the pair of numbers (it has nothing to do with who is a and who is b). Since it is summation, you need to add up the 50 numbers to the maximum. We can enumerate several groups If you enter 100 and 99, you will get 100. If we take two groups of numbers 100 and 1, 99 and 2, then the sum of these two groups of numbers is 199. If we take 100 and 99 in this way, we will get 100 2 and 1, then the sum of these two groups of numbers is 102, so it can be seen clearly that large numbers and large numbers should not meet, so that the final sum can be maximized. Therefore, as long as the largest 50 numbers in 100 natural numbers are from 51 to 100, any two numbers are different groups, so that the final sum of 50 numbers is the maximum of 50 numbers from 51 to 1 00, 51 + 52 + 53 + +100 = 3775. So the answer is: 3775
When a takes any number that Na + 3 is not equal to 0, the formula MA-2 / Na + 3 is a fixed value, where M-N = 6
Let a = 0, MA-2 / NA-3 = - 2 / 3, that is, the fixed value is - 2 / 3
Let a = 1, m-2 / n-3 = - 2 / 3 and M-N = 6
So we have (n + 6-2) / (n + 3) = - 2 / 3
That is, 3 (n + 4) = - 2 (n + 3) gives n = - 18 / 5, M = 12 / 5
For a project, Party A will complete it in 10 days and Party B will complete it in 20 days. The work efficiency of Party C is lower than that of Party A but higher than that of Party B. how many days will it take for three people to complete the project?
If the minimum number of days is required, assume that it will take 10 days to complete
x*（1/10+1/10+1/20）=1
x=4
If an integer part greater than 1 is divided by 167352574 to get the same remainder, then the integer is______ ．
352-167 = 185 = 5 × 37574-352 = 222 = 2 × 3 × 37574-167 = 407 = 37 × 11; so this integer is the common factor of three differences: 37; answer: this integer is 37
A, B, C three people take the train to a certain place, because each of their luggage exceeds the free weight, they need to add a luggage fee. A paid 3 yuan, B paid 5 yuan, C paid 7 yuan. The total weight of the three people's luggage is 90 kg, if these luggage are carried by one person, they need to pay 35 yuan, how much is the weight of C's luggage?
The total weight of three people's luggage is 35 yuan more than that of one person's free weight, but it is divided into three people. The amount of money in excess of the weight is 3 + 5 + 7 = 15 yuan. So if one person's free weight is calculated as overweight weight, the amount of money needed is (35-15) / 2 = 10 yuan. So the ratio of three people's luggage weight is (10 + 3): (10 + 5): (10 + 7) = 13:15:17
Cut off two ropes of different lengths by the same length, and the remaining length ratio is 2:1. If you cut off two ropes again by the same length as last time, the remaining length ratio is 3:1______ ．
Let the length of the original two ropes be x and Y respectively. From the meaning of the question, we can get: (x-a): (Y-A) = 2:1, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; x-a = 2y-2a, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp
The engineering team needs to build a canal: if it is built 8 meters more every day, it can be completed 4 days in advance; if it is built 8 meters less every day, it will be completed 4 days later?
Suppose that the specified time is x days, and the workload of 4 days is: 8 [x-4], then one day can be repaired: 2 [x-4] [2 (x-4) - 8] [x + 8] = x * 2 [x-4] [2x-16] [x + 8] = 2x ^ 2-8x 2x ^ 2 + 16x-16x-128 = 2x ^ 2-8x, 8x = 128, x = 16, that is, the specified time is 16 days, one day can be repaired: 2 * [16-4] = 24, so
That is to say, the equation can only be listed once in one variable
AB and ab are facing each other from a and B. the first encounter is 90km away from a, and the second encounter is 30km away from B. the two encounters are opposite (the second encounter is not overtaking). Find the distance between the two places
The answer is 240, a quadratic equation of one variable can be solved. Now the problem is, there is no way to teach children
That is to say, the equation can only be listed once in one variable
Draw a line diagram to see
The two met for the first time and had a total journey
Among them, line a covers 90 kilometers
Two people meet for the second time, altogether line 3 entire journey
It takes three times as long as a total journey
A should be 90 × 3 = 270 km
It's known that line a has completed one full journey plus 30 kilometers
Here is an equivalent relationship
One full journey plus 30 kilometers equals 270 kilometers
Whole journey: 270-30 = 240 km
Comprehensive formula: 90 × 3-30 = 240 km
The train station starts selling tickets at 8:00, but people have been waiting in line for a long time. Since the first person waiting to buy tickets arrives, the number of people coming every minute is the same. If three windows are opened to sell tickets, no one will queue up at 8:9. If five windows are opened to sell tickets, no one will queue up at 8:5. What is the arrival time of the first person waiting to buy tickets?
Hello, I'm glad to answer for you
If each window is set to sell "1 unit" tickets per minute
Then the sum of the number of people waiting in line before 8:00 and the number of people arriving within 9 minutes = the number of tickets sold in each window per minute × time × number of windows, i.e. 1 × 9 × 3 = 27
The sum of the number of people waiting to buy tickets before 8:00 and the number of people coming 5 minutes is 1 × 5 × 5 = 25
The number of people coming to buy tickets per minute is equal to the difference between the total number of people in 9 minutes and 5 minutes △ the time difference
That is, (27-25) / (9-5) = 0.5 (that is, people who can come to 0.5 units per minute)
The number of people waiting to buy tickets before 8:00 unit = the total number of tickets sold in 9 minutes in 3 windows unit - the number of people coming to buy tickets in 9 minutes unit
27--0.5×9=22.5
Use 22.5 units of people waiting to buy tickets before 8:00 △ 0.5 units per minute = 45 minutes for these people to arrive
Then, the first person arrived at 8:00-45:00 = 7:15
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