Simple algorithm formula of mathematics in grade eight

Simple algorithm formula of mathematics in grade eight

x2-y2=(x+y)(x-y)
(X+y)2=x2+2xy+y2
How to decompose 3x ^ 2-10x ^ 2 + 3x by cross multiplication?
Title, such as
-8x^2+3x=x（-8x+3）
Five cards a b c d e where C can't be placed at both ends how many ways are there
3 * 4 * 3 * 2 * 1 = 72 species
X ^ 2-3x + 2 cross multiplication
x^2-3x+2
=（x-1)(x-2)
1 -2
1 -1
-1*1+(-2)*1=-3
So if (X-2) (x-1) = 0, the solution is 1.2
A permutation and combination problem, seeking mathematical talent
7 choose 5. And you can't choose repeatedly. In the case of pure mongering, what's the probability of all the mistakes?
The answer on the first floor is obviously wrong
Because there are seven answers, the total number is 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040, there is one right answer, 78 wrong answers, 231 wrong answers, and so on, there are 219 wrong answers, so the probability is 219 / 5040 = 0.0434523895238095
The cross multiplication solution of 3x square + 2x-6 = 0
It is not convenient to solve this problem by cross multiplication
3x²+2x-6=0
x² + 2x/3 - 2=0
x² + 2x/3 + 1/9 - 19/9 = 0
(x+1/3)² = 19/9
x+1/3 = ±√19/3
x = -1/3 ±√19/3
How many are the probabilities of a certain two digit combination (5,9) appearing in five digit combinations at the same time? We hope to provide the calculation formula~
There are 332640 possibilities for any combination of five numbers in 11: (11x10x9x8x7) / (5x4x3x2x1);
There are 60480 possibilities for two specific digit combinations (5,9) to appear simultaneously in five digit combinations: (9x8x7) / (3x2x1);
so
The probability of two specific digit combinations (5,9) appearing in five digit combinations at the same time is 60480 / 332640 = 2 / 11
There are five kinds of C (11,5) numbers extracted from 11 numbers - if you can't type this symbol, you'll make do with it
The required number combination contains (5,9), which means that any 3 numbers can be extracted from the other 9 numbers, with a total of C (9,3)
So the probability of two specific digit combinations (5,9) appearing in five digit combinations at the same time = C (9,3) / C (11,5) = 2 / 11... Expansion
There are five kinds of C (11,5) numbers extracted from 11 numbers - if you can't type this symbol, you'll make do with it
The required number combination contains (5,9), which means that any 3 numbers can be extracted from the other 9 numbers, with a total of C (9,3)
So the probability of two specific digit combinations (5,9) appearing in five digit combinations at the same time = C (9,3) / C (11,5) = 2 / 11
There's something wrong with the train of thought upstairs
Any five combinations of 11 numbers should have a (11 5) = 11x10x9x8x7
Since 5 and 9 are fixed, the two numbers are arranged as a (52) = 5X4
In addition, find 3 arbitrary permutations of 9 numbers with a (93) = 9x8x7
The answer is 2 / 11
3x ^ - 4x-4 = 0. Explain how this equation can be solved by cross multiplication. If x is unique before the second power, I will, if not, I will not,
3x²－4x－4=0
32
1 －2
Then:
(3x＋2)(x－2)=0
3x + 2 = 0 or X-2 = 0
The results are as follows
X = - 2 / 3 or x = 2
There is a number before the second power of X, and the method is the same
3 can only be 3 * 1
-4==-4*1=-1*4=-2*2
Cross multiplication:
3 2
1 -2
-2*3+2=-4
So the equation = (3x + 2) (X-2) = 0
3x 2
x -2
3x*(-2)+x*2=-4x
So 3x & # 178; - 4x-4 = 0
(3x+2)(x-2)=0
Ask for help
There are five balls numbered 1, 2, 3, 4 and 5 and five boxes numbered 1, 2, 3, 4 and 5. First put these five balls into five boxes
1. Just one box is empty. How many kinds of casting methods are there?
2. No sunspot is empty, but the number of the ball is not exactly the same as that of the box?
3. There is a ball in the paperless box, and the number of two balls is the same as that of the box. There are several kinds of throwing methods
First question
First, pick out two of the five balls and put them together
It is (5 * 4) / (2 * 1) = 10 kinds
At this time, it was divided into four parts
Arrange the four in full order
4! = 24 species
Then select an empty box
5 kinds
All in all
10 * 24 * 5 = 1200 investment methods
Second question
What is the total arrangement
There are only five kinds of perfect matches
5! - 1 = 119 species
The third question
First, select two matching boxes from five boxes
All in all
(5 * 4) / (2 * 1) = 10 kinds
The other three boxes don't match
That's three boxes
A B C
The placement of the ball can only be
B C A or
C A B
There are only two,
therefore
There are 10 * 2 = 20 kinds of casting methods
1.c(5,2)*p(4,4)*5
2.p(5,5)-1
3.c(5,2)*2
How to solve x quadratic - 3x + 2 = 0 by cross multiplication
1 -2
X
1 -1
（1-2x）X（1-x）=0
x=1/2 x=1
（x-1）*（x-2）=0
X = 1 or x = 2