1. The root of equation 2x (x-4) = 5 (x-4) is ()

2x（x-4）=5（x-4）
2x²-13x+20=0
（2x-5）（x-4）=0
x1=2/5 x2=4

The asymptotic equation of hyperbola X & # 178; - 4Y & # 178; = 4 is a y = ± 2x b y = ± 1 / 2 x C Y = ± 4x d y = ± 1 / 4 X

Choose a
x²-4y²=4
x²/4-y²=1
So a = 2, B = 1
The asymptotic equation of hyperbola is y = ± A / BX
So y = ± 2x

What is the number of real roots of the equation cos x = LG x?

Drawing method ~ draw out cosx and lgx, see how many intersections are how many solutions ~ there should be three intersections
When drawing, pay attention to the two points of lgx, (1,0) and (10,1)

What is the derivative of e ^ - x?

=-e^-x

Y = (x / a) ^ B + (B / x) ^ A + (B / a) ^ x (a > 0, b > 0)

The original question becomes:
y = x^b/a^b + b^a x^(-a) + (b/a)^x   (a>0,b>0)
y' = (b/a^b)x^(b-1) + b^a (-a) x^(-a-1) + (b/a)^x ln(b/a)
   = (b/a^b)x^(b-1) -ab^a / x^(a+1) + (b/a)^x ln(b/a)
Two formulas are used in this problem
y=x^b    y'=bx^(b-1)
y=c^x    y'=c^x lnc   c=b/a   a,b>0

How to derive y = (x + 1) ^ 0.5 - (x-1) ^ 0.5

y'=1/2*(x+1)^(-1/2)-1/2*(x-1)^(-1/2)

Find f (x) = log2 (| X-2 | + | 3x-1 |) domain

X is any real number, because (| X-2 | + | 3x-1 |) > 0, then | X-2 | + | 3x-1 | is not equal to 0, when x takes any value, | X-2 | + | 3x-1 | is not equal to 0, so the definition field is x is any real number
Hope to adopt

Maximum and minimum of cubic function, without derivative
Cubic function (4a ^ 3) - (4a ^ 2a) + (AA ^ 2)
A is a constant, a is a variable, find the maximum and minimum, without derivative function
I use derivative function to find out, just want to see other methods, the best formula is what
Again, derivative method, I will, I want other methods

(4a^3)-(4a^2A)+(aA^2)
=a(4a^2-4aA+A^2)
=a(2a-A)^2
The formula can only come to this point. Finding the minimax and minimax are two concepts,
The concept of function extremum is the point where the derivative is zero, so we must use the method of derivation to find the function extremum,

Let a function have a minimum value when x = 1 and a maximum value 4 when x = - 1, and we know that the derivative of this function has y '= 3x ^ 2 + BX + C, then we can find this function

When x = 1 and x = - 1, the function has extremum, then f '(1) = f' (- 1) = 0
3 + B + C = 0, 3-B + C = 0. B = 0, C = - 3
Then f '(x) = 3x ^ 2-3, and f (x) = x ^ 3-3x + a
According to f (- 1) = 4, - 1 + 3 + a = 4, then a = 2
So the function is f (x) = x ^ 3-3x + 2

I just learned how to find the maximum and minimum of derivative. I don't know how to find X after finding the derivative
f（x）=x^3-12x
f'(x) =3x-12
so what?

Let f '(x) = 0, then x = 4
When x0, then f (x) increases
So, x = 4 is a minimum

1. The root of equation 2x (x-4) = 5 (x-4) is ()