Polynomial X & # 178; - x + 5 minus 3x & # 178; - ah, the result is a.2x & # 178; + 5 B. - 2x-x + 9 C. - 2x & # 178; - x + 1 D. - 2x & # 178; + X + 9

Polynomial X & # 178; - x + 5 minus 3x & # 178; - ah, the result is a.2x & # 178; + 5 B. - 2x-x + 9 C. - 2x & # 178; - x + 1 D. - 2x & # 178; + X + 9

The polynomial X & # 178; - x + 5 subtracts 3x & # 178; - 4, and the result is
B.-2x²-x+9
Given the set a = {- 3x, 2x-1, - 4}, B = {x-5,1-x, 9}, and a intersection B = {9}. (1) find the value of X; (2) find the intersection B
(1)x=-3
(2) If - 3x = 9, a intersects B = {9}
If 2x-1 = 9, then a intersects B = {- 4,9}
It's very simple. Because the intersection is only 9
So x = - 3 or x = 5.
X = - 3.
When x = 5, 1-x = - 4, intersection more - 4, does not meet the conditions.
So x = - 3 - ask: - brother. What about the second question?
If the second power of X is - 2x-3 / 3x-5 = x-3 / A + X + 1 / B, find the value of a and B
Yes
If both sides of the equation are multiplied by (x-3) (x + 1), then 3x-5 = a (x + 1) + B (x-3) = (a + b) x + (a-3b)
Then 3 = a + B - 5 = a-3b
The solution is a = 1, B = 2
It is known that the | m | + 1 power of the cubic power y of (m-2) x is a binomial of degree x and Y. try to find the value of the quadratic power-m + 1 of the algebraic formula 2m
(m-2) x ^ 3 * y ^ (i mi + 1) is a binomial of degree six about X, y
So I m i + 1 = 3 M + 2 or M = - 2
And because m-2 can't be equal to 0,
So m = - 2
2m^2- m+1 = 2*(-2)^2 -(-2)+1 =11
|m|+1=6,m=±5
m=5: (2m)²-m+1=100-5+1=96
m=-5: (2m)²-m+1=100+5+1=106
600-900 words
Mathematics diary 1
Tuesday, June 28
At noon today, I was doing my summer math homework. Unfortunately, I encountered a very difficult problem. I thought about it for a long time, but I didn't come up with a solution. However, this problem is like this:
There is a cuboid, the product of the front and the top two areas is 209 square centimeters, and the length, width and height are prime numbers. Find its volume
When I saw it, I thought: this problem is really difficult! What we know is only the product of the area of two surfaces. We need to know the length, width and height of the volume, but it has no hint. How can we start!
Just as I was scratching my ears, one of my mother's colleagues came. He first taught me to use the idea of equation to solve, but I was not very familiar with this method. So he taught me another method: first list the numbers, and then exclude them one by one. We first listed many numbers according to the requirements of the topic, such as prime numbers of 3, 5, 7, 11, and then we began to exclude them, Then we find that there are only 11 and 19 left. At this time, I think: one of these two numbers is the common edge length on the front side of the cuboid in the question; the other is the front side of the cuboid except the one above
The sum of the edge length (and the length is both prime numbers). So I began to distinguish between the two numbers
Finally, I get the result, which is 374 cubic centimeters. My formula is: 209 = 11 × 19 19 = 2 + 17 11 × 2 × 17 = 374 (cubic centimeters)
As like as two peas, I used the knowledge I learned this semester: decomposition of the quality factor to check the problem.
After solving this problem, I am more happy than anyone else. I also understand a truth: mathematics is full of mysteries, waiting for us to explore
Mathematics diary 2
Saturday, August 6
This evening, I saw a puzzle math problem, the title: 37 students want to cross the river, there is a ferry can only take five empty boats, they all want to cross the river, at least how many times to use this boat?
Careless people tend to ignore the "empty boat", that is, they forget to have a boat, so they can only take four people at a time. In this way, the number of 37 students minus one student in the boat leaves 36 students. 36 divided by 4 equals 9. The last student who went to the other side as a boatman also went ashore 4, so they have to walk at least 9 times
Mathematics diary 3
Tuesday, August 9
How many apples are left in the old pear tree garden in the evening?
I think apple trees are three times as big as pear trees. If you want to fertilize two kinds of trees on the same day, Master Wang should fertilize 20 × 3 apple trees and 20 pear trees every day. In fact, he only fertilizes 50 apple trees every day, 10 of which are poor, and finally 80 of which are poor, Master Wang has been applying fertilizer for 8 days. There are 20 pear trees in a day, and 160 pear trees in 8 days. According to the first condition, we can know that there are 480 apple trees. This is to solve the problem with the idea of hypothesis, so I think the hypothesis method is a good way to solve the problem
Mathematics diary 4
Thursday, August 11
Today, I encountered another math problem, and it took me a lot of effort to solve it. The question is: there are 30 birds in two trees, and four birds fly away from tree B. at this time, tree a flies to tree B with three birds, and the birds in two trees are just equal. How many birds are there in two trees?
As soon as I finished reading the title, I knew that it was a reduction problem, so I used the method of reduction problem to solve it. But when I checked it, I found that it was wrong. So I began to do it again more seriously. I thought that if I lost four, there would be as many as 13. If I reduced tree B, there would be 14. If I reduced tree a, there would be 16. The formula is: (30-4) △ 2 = 13; 13-3 + 4 = 14; 30-14 = 16, There are 14 trees
By solving this problem, I understand that no matter what problem you do, you should be careful. Otherwise, even if you master the method, you will make mistakes
Find it on the Internet!!!!!!!!!
There are many on the table
Today, it's overcast and a little cold, but the hearts of the students are warm, because the teacher said that today is the Provincial Working Committee of Customs to organize several large enterprises to carry out the "love red book bag, care for left behind children" book donation activities. At 10 o'clock, the meeting began. Looking at the bags of books handed over to the students, I counted them in my heart: there are 10 good-looking books in each red book bag. There are 6 left behind children in each class, and there are 13 classes in our school. The total number of books is: 10 × 6 × 13 = 780, so I have a special idea in my heart: to carry out the "circulation of old books" activity in the school. As the saying goes
Today, it's overcast and a little cold, but the hearts of the students are warm, because the teacher said that today is the Provincial Working Committee of Customs to organize several large enterprises to carry out the "love red book bag, care for left behind children" book donation activities. At 10 o'clock, the meeting began. Looking at the bags of books handed over to the students, I counted them in my heart: there are 10 good-looking books in each red book bag. There are 6 left behind children in each class, and there are 13 classes in our school. The total number of books is: 10 × 6 × 13 = 780, so I have a special idea in my heart: to carry out the "circulation of old books" activity in the school. As the saying goes, "it's better to give someone money than words.". Every year, the students will surely get some good books: bought by themselves, bought by family, sent by relatives and friends, etc. As time goes by, the books have been read for a long time and become old. Many of us don't know what to do with so many books. It's a pity to lose them. We don't know where to put them. If all the old books of the students are stored up, on the one hand, the students can read more books, on the other hand, those books that are not used usually have a better place to go, and at the same time, some students can get rid of the bad habit of littering old books. Take our school as an example. There are 474 people in the school. If each person recycles 10 old books on average, there are 474 × 10 = 4740 books. There are 13 classes in our school, one copy for each class, so our class can have 4740 × 131 ≈ 365 books circulated once. These more than 300 books can be circulated in our class for at least one semester. With these books, on the one hand, we can read a lot of good books, many students' old books also have a place to go, on the other hand, we can cultivate students' good habits of diligence and thrift. More importantly, we can save a lot of resources for our country. According to the survey, a 20-year-old tree can make 3000 pieces of A4 paper. If a book takes 100 pages, it needs 100 △ 2 △ 2 = 25 pieces of A4 paper, and the 4740 books can save 4740 × 25 △ 3000 = 39.5 trees. Isn't that a good way to read? Students, let's act quickly. Comments: a book may ignite a child's hope, a book may change a child's life, a good book is a window to the world, a road to success. As the ancients said, "it's better to teach people to fish than to teach them to fish.". Through a real book donation activity, the young writer can associate with a second-hand book circulation activity in the school, which is really original. In this article, the author assumes that every student in the school donates 10 old books to calculate that the whole school will collect 4740 books, and then divide the books into 13 classes on average, each class can see more than 300 books at a time. What a good idea! Let more students read more books, but also cultivate the good habit of diligence and thrift. Put it away
In the evening, I saw a problem in the Olympic book: the apple trees in the orchard are three times as big as the pear trees. Master Wang fertilizes 50 apple trees and 20 pear trees every day. After a few days, all the pear trees are fertilized, but there are still 80 apple trees left. Excuse me: how many apple trees and pear trees are there in the orchard? I'm not frightened by this problem. The difficult problem can arouse my interest. I think the apple tree is three times as big as the pear tree. If we want to fertilize the two kinds of trees on the same day, Master Wang should fertilize "20 × 3" apple trees and 20 pear trees every day. In fact, he only fertilized 50 apple trees every day, 10 of them were poor, and finally the total difference was
In the evening, I saw a problem in the Olympic book: the apple trees in the orchard are three times as big as the pear trees. Master Wang fertilizes 50 apple trees and 20 pear trees every day. After a few days, all the pear trees are fertilized, but there are still 80 apple trees left. Excuse me: how many apple trees and pear trees are there in the orchard? I'm not frightened by this problem. The difficult problem can arouse my interest. I think the apple tree is three times as big as the pear tree. If we want to fertilize the two kinds of trees on the same day, Master Wang should fertilize "20 × 3" apple trees and 20 pear trees every day. In fact, he only fertilized 50 apple trees every day, 10 of them were poor, and finally 80 of them were poor. It can be seen from here that Master Wang has been fertilizing for 8 days. There are 20 pear trees in one day, and 160 in eight days. According to the first condition, we can know that there are 480 apple trees. This is to use the idea of hypothesis to solve problems, so I think the hypothesis method is really a good way to solve problems. Put it away
How to use the graph of the first order function y = 3x + 12 to find the solution set of the inequality 3x + 12 > 0
The intersection of the function image and the X axis is the point where the function value y = 0, that is, let
3X+12=0
The image above the x-axis is the part of Y 〉 0, that is, 3x + 12 > 0. Obviously, the range of abscissa x corresponding to this part is the solution set of inequality
The x value of the image with Y > 0 is
The point above the line y = 3x + 12 satisfies the inequality 3x + 12 > 0
When y is greater than 0, the value range of X is the solution set of X
Two intersection points of the image of quadratic function y = AX2 + BX + C and x-axis, the intersection point is a, B, and the vertex is C. find s △ ABC
S△ABC=√(b^2-4ac)*|c/a|/2
Molecule: △ under △ radical
Denominator: the absolute value of 8A * a
[under radical (b ^ 2-4ac)] * (C-B ^ 2 / 4A) / 2A
1/2(b^2/a^2-4c/a)^1/2(b^2/2a-b^3/2a+c)
Suppose a