What is the value range of the independent variable of the inverse proportion function? For example: y = under root (X-2) / (x + 3) Value range of independent variable Is this an inverse scale function?

This is a compound function, there are inverse proportional function model, there are also radical function
The range of the independent variable of this function is the domain of definition of the function: the denominator of the fraction is not zero and the radical is greater than or equal to zero

The value range of independent variable of inverse scale function y = 10 / X

X is not equal to 0
Because x is the denominator

If the image of function y = KX passes through the point (- 1,3), then the value of K is ()

x=-1,y=3
So 3 = - K
k=-3

X ^ 3 + x ^ 2 + X + 1 = 0, then x ^ 97 + x ^ 98 + x ^ 99 +. X ^ 103=

Put forward the common factor method X ^ 97 + x ^ 98 + x ^ 99 + x ^ 100 + x ^ 101 + x ^ 102 + x ^ 103 = x ^ 97 (1 + X + x ^ 2 + x ^ 3 + x ^ 4 + x ^ 5 + x ^ 6) because 1 + X + x ^ 2 + x ^ 3 = 0, so the original formula = x ^ 97 (x ^ 4 + x ^ 5 + x ^ 6) = x ^ 100 (x + x ^ 2 + x ^ 3) from 1 + X + x ^ 2 + x ^ 3 = 0 → x + x ^ 2 + x ^ 3 = - 1, the original formula = - x ^ 100

98-x, we can get (98-x-9) 4 = X-9 and solve the equation

(98-X-9）4=X-9
（89-x）×4=x-9
356-4x=x-9
x+4x=356+9
5x=365
x=73
98-73=25

4: X = x: 9 to solve the equation

4:X=X:9
X²=4*9
X²=36
X=±6

The solution equation: x + (9-x) = 9

X = any number, study hard and make progress every day

98 out of 99 × 100-98 out of 99 is easy to calculate! X + 3x out of 4 = 15 out of 9

= 98 / 99 × (1 + 99) - 98 / 99
98-99
=98

Design a calculation 1 + 3 + 5 + 7 + The algorithm of 99? How to calculate

= (1+99) + (2+98) + ...+ (49+51) + 50
= 100 * 49 +50
=4950

Design a calculation 1 + 1 / 2 + 2 / 3 + 3 / 4 + +99 / 100 algorithm, and draw the program block diagram
Design a calculation 1 + 1 / 2 + 2 / 3 + 3 / 4 + +99 / 100 algorithm, and draw the program block diagram

REM language: Visual Basic 6.0sub main() & nbsp; & nbsp; start & nbsp; s = 1: I = 2 & nbsp; do & nbsp; & nbsp; & nbsp; s = S + (I - 1) / I & nbsp; & nbsp; if not (I & lt; 100) then exit do & nbsp; & nbsp; I = I + 1 & nbsp; loop & nbsp; msgbox s, vbinformation, & quot; output & quot; End Sub & nbsp; & nbsp; end

What is the value range of the independent variable of the inverse proportion function? For example: y = under root (X-2) / (x + 3) Value range of independent variable Is this an inverse scale function?