Given the power function of the image through the point (8,1 / 4), find the value of F (27)

Given the power function of the image through the point (8,1 / 4), find the value of F (27)

Let f (x) = x ^ A
The image of power function passes through the point (8,1 / 4),
That is, 8 ^ a = 1 / 4
That is, 2 ^ 3A = 2 ^ (- 2)
That is 3A = - 2
That is a = - 2 / 3
So f (x) = x ^ (- 2 / 3)
so
f(27)=27^（-2/3)=(3^3)^（-2/3)
=3^(-2)
=1/3^2
=1/9,

The sum of us is 17, the product of us is 52, a composite number and a prime number

13 and 4. It's very simple

9(x-5y)²-49(x+5y)²

9(x-5y)²-49(x+5y)²=[3(x-5y)]²-[7(x+5y)]²=[3(x-5y)+7(x+5y)][3(x-5y)-7(x+5y)]=(3x-15y+7x+35y)(3x-15y-7x-35y)=(10x-20y)(-4x-50y)=-20(x-2y)(2x+25y)

x^2-4xy+5y^2+6y+9=0,x+y=

x^2-4xy+5y^2+6y+9
=x²-4xy+4y²+y²+6y+9
=（x-2y）²+（y+3）²
=0
∴x-2y=0 y+3=0
∴x=-6 y=-3
∴x+y=（-6）+（-3）=-9

- X & # 178; + 4Y & # 178; + 4xy factorization

﹣x²＋4y²+4xy
=4y²+4xy +x²﹣2 x²
=(2y+x)-2x²
=(2y+x+√2x)(2y+x-√2x)

The two trains run on two parallel tracks, of which the length of the express train is 100 meters and the length of the local train is 150 meters. When the two trains run in opposite directions, it takes 5 seconds for the express train to pass a window of the local train (from the point where the front of the express train arrives at the window to the point where the rear of the train leaves). (1) find the sum of the speeds of the two trains and the time when the two trains run in opposite directions (2) if two trains are traveling in the same direction, the speed of the slow train is not less than 8 m / s, and the fast train chases the slow train from behind, then how many seconds is the time from the front of the fast train catching up with the back of the slow train to the back of the fast train leaving the slow train?

(1) Let the speed of the fast train and the slow train be x m / s and Y M / s respectively. According to the meaning of the title, we get x + y = 1005 = 20, that is, the sum of the speed of the two trains is 20 m / s. let the slow train take T1 seconds to pass a window of the fast train. According to the meaning of the title, we get x + y = 150t1, ｜ T1 = 150x + y = 15020 = 7.5

What is coefficient determinant
Baidu does not arrive. So come to ask a question
Please make it easy to understand~
Given a matrix, how to find its coefficient determinant?

Generally speaking, for a system of linear equations, it has a coefficient matrix, which is composed of the coefficients of the unknowns. The determinant of this matrix is the coefficient determinant. For example, for the system of equations, a1x1 + a2x2 + a3x3 = 0b1x1 + b2x2 + b3x3 = 0c1x1 + c2x2 + c3x3 = 0, its coefficient matrix is [A1, A2

-Factorization of 4x2 (2x - 3Y) 2
The 2 after each item is square. Sorry, there is no wealth value. Urgent!

-4x2+ (2x -3y)2
=(2x-3y)²-4x²
=(2x-3y+2x)(2x-3y-2x)
=-3y(4x-3y)

Given the set u = {2.4.1-a}, a = {- 1}, CUA = {2, a & sup2; - 1 + 2}, then the real number a=
U = {2,4,1-a}, I'm wrong=

Because CUA = (2, A2-1 + 2), A2-1 + 2 = 4 and not equal to - 1, so 1-A is not equal to - 1, A2-1 + 2 = 4, so a = plus minus sign 3

It is proved that the function f (x) = x (1 / (2 ^ x-1) + 1 / 2)) + a (where a is constant) is even function

Certification:
The domain x ≠ 0 is symmetric about the origin
f(x)=x(1/(2^x-1)+1/2))+a
f(-x)=-x*{1/[2^(-x)-1]+1/2)}+a
=-x*{2^x/(1-2^x)+1/2}+a
=-x[(2^x-1+1)/(1-2^x)+1/2]+a
=-x[-1+1/(1-2^x)+1/2]+a
=-x[1/(1-2^x)-1/2]+a
=x*[1/(2^x-1)+1/2]+a
=f(x)
So f (x) is an even function