If a is a real number not equal to 1, we call 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-2 = - 1, and the difference reciprocal of - 1 is 1 / 1 - (- If a is not equal to 1, we call 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-2 = - 1, and the difference reciprocal of - 1 is 1 / 1 - (- 1) = 1 / 2. If A1 = - 1 / 3, A2 is the difference reciprocal of A1, A3 is the difference reciprocal of A2, A4 is the difference reciprocal of A3, and so on, then a2013=_______ .

If a is a real number not equal to 1, we call 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-2 = - 1, and the difference reciprocal of - 1 is 1 / 1 - (- If a is not equal to 1, we call 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-2 = - 1, and the difference reciprocal of - 1 is 1 / 1 - (- 1) = 1 / 2. If A1 = - 1 / 3, A2 is the difference reciprocal of A1, A3 is the difference reciprocal of A2, A4 is the difference reciprocal of A3, and so on, then a2013=_______ .

The solution to this problem should be periodic sequence,
a1=-1/3
a2=1/（1-（-1/3））=3/4
a3=1/（1-a2）=1/(1-3/4)=4
a4=1/（1-a3）=1/(1-4)=-1/3
a5=1/(1-a4)=1/(1-(-1/3))=3/4
.
The stock cycle is 3
And 2013 = 671 × 3
So a2013 = A1 = - 1 / 3
How to draw function image when it is not equal to zero?
The value range of X is not equal to 0.
And then it's gone
Images just learned today
But did not learn the case of X ≠ 0
I don't know if the curve goes through the origin --
Just hollow the point of intersection with the y-axis
What doesn't equal 0? Is it X or Y or the square of B minus 4ac?
If the real numbers a and B are opposite to each other, (AB is not equal to 0), C and D are reciprocal to each other, and the absolute value of M is 3, find a + B of ab+
If the real numbers a and B are opposite to each other, (AB is not equal to 0), C and D are reciprocal to each other, and the absolute value of M is 3, find the value of a + B + CD divided by (2 + 3M + m square)
1 / 20 when m = 3
When m = - 3, it is 1 / 2
How many grams is 0.5L?
If you draw an image with function y equal to minus 2 / x, you should pay attention to that the value of X can't be 0, right
yes.
More Than This:
So the image with y = - 2 / X is divided into two parts,
That is, the function image has two branches
If (2a-1) and (2a + 1) are reciprocal, then the real number a is equal to
(2a-1) and (2a + 1) are reciprocal
Then (2a-1) * (2a + 1) = 1
That is 4A ^ 2-1 = 1
So a ^ 2 = 1 / 2
So a = - 2 / 2
A = root 2 of 2
Solution equation (2A - 1) = 1 / (2a + 1)
A = positive and negative root sign 2 / 2
Two mathematics are reciprocal to each other, so the multiplication of two mathematics is equal to 1
So 4 * a ^ 2-1 = 1
If a > 0 and a is not equal to 1, then the image of function y = a ^ (x-1) - 1 must pass through the fixed point------
When x = 1, y = a ^ 0-1 = 1-1 = 0
So, the function passes the fixed point (1,0)
(1,0)
If real numbers a and B satisfy B = √ (a-5) + √ (5-a) + 4, and C is the number whose reciprocal is equal to itself, try to judge the root of the equation AX & # 178; + BX + C = 0 (a ≠ 0) about X
If a =  B =  - 4,  a =  - 4,  - B =  - 4,  - 5,  - 4,  - 4,  - 4,  - 4,  - 4,  - 4,  - 4,  - 4,  - 5,  - 4,  - 4,  - 5,  - 4,  - 4,  - 4,  - 4,  - 5,  - 4,      - 4,
solution
A-5 ≥ 0 and 5-a ≤ 0
∴a=5∴b=4
∵ C is the number of the reciprocal itself
∴c=±1
When C = 1, the equation is 5x & # 178; + 4x + 1 = 0
△=4²-4×5=16-200
Two unequal real roots of the equation
a. B satisfies B = √ (a-5) + √ (5-a) + 4,
Because a-5 > = 0,5-a > = 0
So there is a > = 5, a = 0 and 5-a > = 0, so a = 5 and B = 4
b^2-4ac=16-4*5*1=-40
So the equation AX & # 178; + BX + C = 0 (a ≠ 0) has two different solutions
Given the function f (x) = a ^ (x-1), (x > = 0), the image passes through the point (2,1 / 2), where a > 0 and a is not equal to 1
1. Finding the value of a
2. Finding the range of function y = f (x) (x > = 0)
① If ^ (2,2) = X-1,
a=1/2
f(x)=1/2^（x-1）
②∵ x>=0 x-1≥-1
∴ 1/2^（x-1）≤1/2^(-1)=2
The range of function y = f (x) (x > = 0)
0＜y≤2
1. The image of ∵ f (x) passes through the point (2,1 / 2)
∴a^(2-1)=1/2
The solution is a = 1 / 2
2. ∵ y = (1 / 2) ^ x decreases monotonically on R in the range of (0, + ∞)
The maximum value of F (x) = (1 / 2) ^ (x-1) (x > = 0) is f (0) = 2
The range of y = f (x) (x > = 0) is (0,2]
That's all. I wish you progress in your study.
1 / 2 = a ^ (2-1) and a = 1 / 2
⑵f(x)=(1/2)^(x-1)
When x ≥ 0, X-1 ≥ - 1
∴f(x)≤2
That is, the value range is (0,2]
Note that the final range should be written as a set or interval, and distinguish between open and closed intervals
(1) A ^ 1 = 1 / 2, so a = 1 / 2
(2) F (x) = (1 / 2) ^ (x-1), because x > = 0, so X-1 > = - 1, and the function f (x) is a monotone decreasing function, so f (x)
Why do we still call two real numbers reciprocal when the product of two real numbers (a + b) and (a-b) is 1
Wait, I read the wrong topic! Sweat first, no matter it's useful or not, thank you first.
The original title should be: when the product of (B under a + radical) and (B under a - radical) is 1, (B under a + radical) and (B under a - radical) are reciprocal.
Here to answer the netizen apology——
Because the product of two reciprocal numbers must be 1, so I think so
We know that if the product of two rational numbers is 1, then the two rational numbers are reciprocal to each other+
b) When the product of (a-b) and (a-b) is 1, we still call these two real numbers reciprocal to each other
Because... The teacher said that two reciprocal x must be one
I don't know the root, but I haven't learned it yet Thank you. Please~
If the function f (x) = a ^ (2-x) - 1 (A & gt; 0, and a is not equal to 1), the image passes through the fixed point
Function over fixed point means: no matter how a changes, f (x) must be a fixed value. So when exponent 2-x = 0, that is, x = 2, no matter what number a is, a ^ (2-x) = 1 is a fixed value, then f (x) = a ^ (2-x) - 1 = 0 is also a fixed value. Function over fixed point (2,0)