The operation of high school factorial I know n! It's multiplying n by 1 What I want to know is factorial, factorial operation and quick calculation For example: 1, M is less than or equal to N, then / = n (n-1) How to push (M + 1)? 2. Seven people stand in a row. How many kinds of arrangement are there in the first row? Why is A6 (6) the case of a standing at the top of the row when using the exclusion method

The operation of high school factorial I know n! It's multiplying n by 1 What I want to know is factorial, factorial operation and quick calculation For example: 1, M is less than or equal to N, then / = n (n-1) How to push (M + 1)? 2. Seven people stand in a row. How many kinds of arrangement are there in the first row? Why is A6 (6) the case of a standing at the top of the row when using the exclusion method

1),n!/ m!=1×2×3×…… × m×（m+1）×…… （n-1） ×n/1×2×3×…… ×m=（m+1）×（m+2）×…… ×（n-1）×n.
I didn't understand the second question
Eighty-eight
Calculate and output the factorial of 9!
JX=1
N=1
DO WHILE[ ]
JX=JX*N
[]
ENDDO
'9!='+'1*2*3*4*5*6*7*8*9='+[ ]
JX=1
N=1
DO WHILE N
The factorial of 9 is 362880
A=1
b="1"
for i=2 to 9
a=a*i
b=b & "*" & i
next
print "9!=" & b & "=" & a
Results: 9! = 1 * 2 * 3 *.... * 9 = 362880
If f (x) = (m-1) x2 + (m-2) x + (m2-7m + 12) is even function, then the value of M is ()
A. 1B. 2C. 3D. 4
∵ function f (x) = (m-1) x2 + (m-2) x + (m2-7m + 12) is even function, ∵ f (- x) = f (x), ∵ m-1) X2 - (m-2) x + (m2-7m + 12) = (m-1) x2 + (m-2) x + (m2-7m + 12), ∵ m-2 = 0, M = 2, so B
The domain of function f (x) = √ Log1 / 2 (x-1)
f(x)=√log1/2(x-1)
X needs to satisfy: X-1 > 0, log (1 / 2) (x-1) ≥ 0
Ψ x ＞ 1 and 0 ＜ X-1 ≤ 1
∴1＜x≤2
So the domain is (1,2]
X>1
Given that f (x) = (M2-1) x2 + (m-1) x + N + 2, when m, n are values, f (x) is an odd function
The so-called odd function is f (- x) = - f (x). For power function, when the coefficient of even power term (even power term of x) is 0 and the coefficient of odd power term is not 0, the function is odd
m2-1=0
m-1≠0
N + 2 = 0
The solution is m = 1 or M = - 1
And m ≠ 1
And N = - 2
In conclusion, M = - 1 and N = - 2 are the results
M = - 1, n = - 2
The domain of the function f (x) = Log1 / 2 1 / 1-x is
It's clear that the answer is wrong
X=0
F (x) = Log1 / 2 (1) = 0
Unless it's in the denominator
The domain is X
If the function f (x) = (m-1) x2 + (m-2) x + M2 is even, then the value of M is
F (x) = (m-1) x & sup2; + (m-2) x + M & sup2; is even function
Then f (x) = f (- x)
f(-x)=(m-1)x²-(m-2)x+m²
That is, (m-1) x & sup2; + (m-2) x + M & sup2; =) = (m-1) x & sup2; - (m-2) x + M & sup2;
So m-2 = - (m-2)
The solution is m = 2
∵ f (x) is an even function
Even functions should be symmetric about the y-axis
∴m-2=0 m=2
Mathematics of grade one in Senior High School: seeking the domain of definition, monotone interval and range of values 1: y = 1 / 2 ^ 1 / x 2: y = 3 / 7 ^ (x ^ 2-2x) 3: y = under the root sign (4 ^ X-2 ^ x) 4: y = (2 ^ x + 1)
1 is an exponential function with 1 / 2 as the base and 1 / X as the index. This is nothing special, as long as the denominator is not zero, because the index can be anything. So the domain of definition is x ≠ 0, (- ∞, 0) ∪ (0, + ∞). Because u = 1 / X is a decreasing function, and the exponential function with 1 / 2 as the base is also decreasing
Let f (x) be a decreasing function defined on (- 2,2) and satisfy the following conditions: F (- x) = - f (x), and f (m-1) + F (2m-1) > 0, then the value range of real number m is obtained
The inequality f (m-1) + F (2m-1) > 0, that is, f (m-1) > F (2m-1), ∵ f (- x) = - f (x), we can get - f (2m-1) = f (- 2m + 1) ∵ the original inequality is transformed into f (m-1) > F (- 2m + 1) and ∵ f (x) is a decreasing function defined on (- 2,2), ∵ - 2 < M-1 < - 2m + 1 < 2, the solution is - 12
The range of function y = x / 1 + X (x ≠ - 1) is
y=(x+1-1)/(x+1)
=(x+1)/(x+1)-1/(x+1)
=1-1/(x+1)
1/(x+1)≠0
So the range (- ∞, 1) ∪ (1, + ∞)
Substituting - 1, the range is y, not - 2
y=(x+1-1)/x+1=1-1/(x+1),
Therefore, the range is a set of all real numbers without 1.