It is known that the domain of definition of functions f (x) and G (x) is r, where f (x) is odd, G (x) is even, and f (x) + G (x) = 1 / (the square of x-x + 1) Finding the analytic expressions of F (x), G (x)

It is known that the domain of definition of functions f (x) and G (x) is r, where f (x) is odd, G (x) is even, and f (x) + G (x) = 1 / (the square of x-x + 1) Finding the analytic expressions of F (x), G (x)


Let H (x) = f (x) + G (x)
h(x)+h(-x)= f(x)+g(x)+f(-x)+g(-x)=2g(x)
h(x)-h(-x)= f(x)+g(x)-f(-x)-g(-x)=2f(x)
So it's very simple: H (x) + H (- x) divided by 2 to get g (x)
Divide H (x) - H (- x) by two to get f (x)



It is known that f (x) is an odd function, G (x) is an even function, f (x) + G (x) = LG (x + 1)
1. Find f (x) and G (x). 2. If G (x) is less than a in the domain of definition, find the range of A


(1) It is easy to know that f (x) + F (- x) = 0. G (x) = g (- x) (- 1 < x < 1). F (x) + G (x) = ㏑ (x + 1), f (- x) + G (- x) = ㏑ (1-x & sup2;). By adding the two formulas, G (x) = ㏑ √ (1-x & sup2;). By subtracting the two formulas, f (x) = ㏑ √ [(1 + x) / (1-x)]. (- 1 < x < 1). (2) g (x) = ㏑ √ (1-x & sup2;) < a, (- 1 < x < 1)



The product of 19 and () is prime


A:
The product of 19 and (1) is prime
It must not be multiplied by a number greater than 1, otherwise it is not prime
Because the result of multiplication is prime = 19 * X
Then the prime factor can only be 19 and 1



Some problems of function image in junior high school mathematics
1. Given the points M1 (- 1, - 2), M2 (- 2, - 1), m3 (1,2), M4 (2,1), which are not on the graph of function y = x + 1, there are -
2. If the point (- 2,3) exists on the image of function y = 2x + 2m, then M = --
3. If the point (- 6, a under the root sign) is on the image of the function y = 2x & # 178; - 1, then a = --
4. If the image of a function is in one or two quadrants, then the value of the function ()
A. All integers B. all negative numbers C. all non negative numbers D. can be positive or negative or 0
5. Given that point (3,6) is on the graph of function y = K / X (K ≠ 0), then the following points are ()
A.(-3,6) B.(2,9) C.(2.-9) D.(3.-6)


1. Given the points M1 (- 1, - 2), M2 (- 2, - 1), m3 (1,2), M4 (2,1), which are not on the image of function y = x + 1 are: M1, M4.2, if the point (- 2,3) exists on the image of function y = 2x + 2m, then M = 3.5.3, if the point (- 6 under the root sign, a) exists on the image of function y = 2x & # 178; - 1, then a = 11.4, if a function



As shown in the figure, it is known that the line AB = 10, AC = BD = 2, and point P is a moving point on CD. Take AP and Pb as edges, make squares APEF and phkb up and down respectively. Let the intersection points of the square diagonals be O1 and O2 respectively. When point P moves from point C to point D, the length of the motion path of point G in line O1O2 is___ .


As shown in the figure: when P moves to point C and point D, it is concluded that the moving route of point G is a straight line. By using the properties of square, that is, the length of the moving path of point G in line O1O2 is the length of o2o ″, ∵ line AB = 10, AC = BD = 2, when p coincides with C, take AP and Pb as edges, make square APEF and phkb upward and downward, ∵ AP = 2, BP = 8, then o1p = 2, o2p = 42, ∵ o2p = o2b = 42, when p ′ coincides with D, Then p ′ B = 2, AP ′ = 8, O ′ P ′ = 42, O ″ P ′ = 2, H ′ o ″ = Bo ″ = 2, o2o ″ = 42-2 = 32



The minimum natural number of digits 1 or 0 in 10 system and divisible by 225
Such as the title
The answer is 100 * (10 Λ 9-1) / 9


225 = 15 * 15 = 9 * 25
Because this number only contains the number 0 or 1, it can be divided by 225, that is, it can be divided by 9 and 25. According to the characteristics of the number that can be divided by 9 (it should be the sum of all numbers that can be divided by 9)
Therefore, the number contains at least nine ones,
And because it's divisible by 25, it's a multiple of 100
That is, the number is equal to 11100 = 100 * (10 ^ 9-1) / 9



Let the random variables XY be independent of each other and obey the uniform distribution of (0.1)


FZ (z) = ∫ (∞→ + ∞) FX (x) FY (z-x) DX (1) Z < 0fz (z) = ∫ (∞→ + ∞) FX (x) FY (z-x) DX = 0 (2) 0 ≤ Z < 1fz (z) = ∫ (0 → Z) 1.1dx = Z (3) 1 ≤ Z < 2fz (z) = ∫ (0 → Z-1) 1.0dx + ∫ (Z-1 → 1) 1.1dx = 2-z (4) Z ≥ 2, FZ (z) = ∫ (∞→ + ∞) FX (x) FY (z-x) = ∫ (0 → Z-1) 1.0dx + ∫ (Z-1 → 1) 1.1dx = 2-z (4) Z ≥ 2



Calculation of sin (PI / 7) + ln3-e ^ J * 3 by MATLAB


sin(pi/7)+log(3)-exp(complex(0,3))
Or change the power of e into trigonometric function
sin(pi/7)+log(3)-cos(3)-complex(0,sin(3))
Complex (0,3) stands for imaginary number 0 + J3



Positive 3, negative 5, negative 11, and positive 7 make the result equal to 24 for mixed operation (write one for each time only)


[(+7)-(+3)]×[(-5)-(-11)]
=4×6
=24



A parallelogram has an area of 36 square meters and a bottom of 9 meters. There is a triangle with the same area and bottom. What is the height of the triangle?


36 × 2 △ 9 = 8 (m)