3A ^ 2B ^ 2-4ab + 1 factorization factor

3A ^ 2B ^ 2-4ab + 1 factorization factor

3a^2b^2-4ab+1
=(3ab-1)(ab-1)

Calculation: 3 / 2B + 4 / 3A ^ 2 + 3 / 4AB
Fractional addition and subtraction. Detailed process

3a^2b-7ab^2=-4ab,

It's not right
These two a and B times are not the same
So it's not the same kind
So the result is 3A ^ 2b-7ab ^ 2, which can't be simplified

The parametric equation of the displacement from the fixed point m to the moving point P is t for a straight line passing through points (1,5) and with an inclination angle of 60 degrees

The straight line passes through points (1,5) and the inclination angle is 60 degrees,
Then cos60 & # 186; = 1 / 2, sin60 & # 186; = √ 3 / 2
So the standard form of the parametric equation of a line is
x=1+1/2·t
Y = 5 + √ 3 / 2 · t (t is the parameter)

What are the characteristics of the definition field of the function with parity?

Odd functions are symmetric about the origin, even functions are symmetric about the Y axis

What are the characteristics of the definition field of odd even function? What are the conditions for the function to have parity?

All domains must be symmetric about the origin
Necessary and insufficient conditions

Domain and function parity
In advanced mathematics textbooks, it is mainly based on the definition to judge whether a function is odd or even, as if it has little to do with the domain of definition?

In general, for function f (x) (1) if f (- x) = - f (x) exists for any X in the function domain, then function f (x) is called odd function. (2) if f (- x) = f (x) exists for any X in the function domain, then function f (x) is called even function. (3) if f (- x) = - f (x) exists for any X in the function domain

Must the definition field of odd even function be r?

Not necessarily, there is a definition of odd and even functions, whose domain is symmetrical about the origin, so it is not necessarily R. ① the odd and even properties are the overall properties of functions. ② the domain of odd and even functions must be symmetrical about the origin for the whole domain. If the domain of a function is not symmetrical about the origin, then the function must not be odd (...)

It is known that f (x) is an odd function defined on R, and the function f (x) monotonically decreases on [0,1], and satisfies f (2-x) = f (x). If the equation f (x) = - 1 has real roots on [0,1], the sum of the roots of all real numbers in the interval [- 1,3] is obtained

F (x) is an odd function defined on R. the image is symmetric about the origin,
And the function f (x) decreases monotonically on [0,1], then f (x) decreases monotonically on [- 1,1),
If f (2-x) = f (x), then the f (x) image is symmetric with respect to the line x = 1 (2-x and the corresponding function value of X are equal, no matter what the value of X is, x + (2-x) = 2,
[x+(2-x)]/2=1,
If f (x) decreases monotonically on [- 1,1], then f (x) increases monotonically on [1,3],
F (x) = - 1 has real roots on [0,1], the line y = - 1 and f (x) graph have an intersection on [0,1], and each has an intersection on [- 11) and [1,3]. The line y = - 1 and f (x) graph have two intersections on [-, 3], and these two intersections are symmetric with respect to the line x = 1. The sum of abscissa of intersection points is 2, that is, the sum of roots of all real numbers of F (x) = - 1 on the interval [- 1,3] is 2

The function f (x) = (2 ^ x) / (4 ^ x + 1) defined on (0,1) is known. Proof: the function f (x) is monotonically decreasing on (0,1)

Let 0 ＜ x1 ＜ x2 ＜ 1F (x2) - f (x1) = 2 ^ (x2) / [4 ^ (x2) + 1] - 2 ^ (x1) / [4 ^ (x1) + 1] = [2 ^ (x2) × 4 ^ (x1) + 2 ^ (x2) - 2 ^ (x1) × 4 ^ (x2) - 2 ^ (x1)] / {[4 ^ (x2) + 1] × [4 ^ (x1) + 1]} = {[2 ^ (x2) × 4 ^ (x1) - 2 ^ (x1)] + [2 ^ (x2) - 2 ^ (x1)]