Find a math problem. Urgent! One / 2 + one half / 3 + one third / 4 +99 / 100 =? according to 1 / 2 = 1 / 2 - 1 / 2 Half / 3 = half - one third

Find a math problem. Urgent! One / 2 + one half / 3 + one third / 4 +99 / 100 =? according to 1 / 2 = 1 / 2 - 1 / 2 Half / 3 = half - one third

According to the law, there are
1/1-1/2+1/2-1/3+1/3-…… +1/99-1/100=1/1-1/100=99/100
What grade is it? It's easy to find it by talking about it
99/100
A mathematical problem
How much do you get if you subtract 4 out of 8 and 30 by 18 out of 2 and 30?
5 and 16 / 30
8 and 30 4 minus 2 and 30 18
＝244/30-78/30
＝83/15
About a math problem?
The total score is 100 points, divided into 6 parts, each of which has an odd number of 6; for example, 6, 16, 26, 36 and so on add up to 100. How do you work out?
Impossible, 6 copies, each mantissa is 6, then the mantissa of the total score is 6 (6 * 6 = 36), it can't be 100
Unless there's a negative score
The domain of the inverse function of the function f (x) = [3 Λ (x-1)] + 2 is_____ .
The domain of its inverse function is the domain of its original function
3 ^ (x-1) is to shift 3 ^ x to the right and Y remains unchanged
∴3^（x-1）＞0
∴f（x）＞2
So the domain of inverse function is {x | x ＞ 2}
The domain of inverse function is the domain of original function, 3 ^ (x-1) > 0, 3 ^ (x-1) + 2 > 2, so (2, + infinity)
(2, positive infinity)
The function f (x) = X3 + SiNx + 1 (x belongs to R). If f (a) = 2, then the value of F (- a) is
Since x ^ 3 and SiNx are odd functions, f (x) + F (- x) = 2, so f (a) + F (- a) = 2, so f (- a) = 2-F (a) = 0
When the inverse function of y = f (x) is known, f ^ - 1 (x) = - 1 + 2 ^ x, X ∈ R (1) find the analytic expression of the function f (x), and write the domain of definition (2) if G (x) = 2log2 (2X + 4),
y=f^-1(x)=-1+2^x
2^x=y+1
x=log2[y+1]
The original function is
y=log2[x+1]
x+1>0
x>-1
therefore
The domain is (- 1, positive infinity)
An original function of FX is SiNx, so we can find the integral of F'x
If a primitive function of F (x) is SiNx, then f (x) should be (SiNx) '= cosx
So f '(x) = (cosx)' = - SiNx, then its integral should be: cosx + C, where C is a constant
Y = 3-2x / X-1 (x ≠ 1), inverse function
There should be the process of solving the problem and the reasons.
Really do not understand, seek guidance!!!
y=(3-2x)/(x-1)
y(x-1)=3-2x
yx-y=3-2x
yx+2x=3+y
(y+2)x=3+y
x=(3+y)/(y+2)
y≠-2
therefore
The inverse function is
y=(x+3)/(x+2)
The domain is {x | x ∈ R and X ≠ - 2}
Given the function f (x) = LG (kx-1) / (x-1), (k belongs to R and k > 0), if the function f (x) increases monotonically on [10, positive infinity], the value range of K is obtained
F (x) = LG (kx-1) / (x-1), = LG (kx-1) - LG (x-1) first of all, K is not equal to 1 / 10, because if k = 1 / 10, then (kx-1) / (x-1) = 0 will appear when x = 10. Because K > 0, it is unnecessary to consider that K is not equal to 1. If k = 1, then the true number (kx-1) / (x-1) = 1 becomes a function that neither increases nor decreases. From y = LGN, we can see that the logarithm LGN is an increasing function in its domain, so if K
What is the inverse function of y = LG (4-x) / (x-3)
The inverse solution can replace y with X, y = (4 + 3 * 10x) / (1 + 10x)
10x is the x power of 10
It can be seen from the original formula that (4-x) / (x-3) > 0
Three
Inverse solution, y = (4 + 3 * 10x) / (1 + 10x)
10x is the x power of 10