Function domain y = LG (x ^ 2 + 3x-28)

x^2 +3x-28>0
(x-4)(x+7)>0
x> 4 or think

Given that the function f (x) defined on R satisfies f (2) = 3 and f '(x) - 1 < 0, then the solution set of the inequality f (x ^ 2) < x ^ 2 + 1 is_______

Let g (x) = f (x) - X
Derivation of G (x)
Derivative of G (x) = f (x) '- 1

It is known that the function f (x) defined on R satisfies f (1) = 2, f '(x)

Let f (x) = f (x) - x, f '(x) = f' (x) - 1

The square of an even number must be even, and vice versa. If an even number is a complete square, its square root must also be even. How to prove it

Proof: to the contrary
Let 2m = n & sup2;, suppose n is odd, let n = 2T + 1;
Then 2m = (2t + 1) & sup2;,
4 (T & sup2; + T) = 2m-1
Even numbers on the right and odd numbers on the left are obviously not tenable, so the hypothesis is not tenable;
So n must be even

Difficult problems in quadratic equation of one variable
It is known that the quadratic equation x ^ 2 + BX + C = x has two real roots x 1, x 2, and satisfies x 12 (B + 2C)
(Note: note that x follows the equal sign, ^ 2 represents the square, X1 and X2 are marked with small 1 or small 2 below x respectively.)

The original formula can be reduced to x ^ 2 + (B-1) x + C = 0
Because the original equation has two real roots, it has X1 = [- (B-1) - sqrt ((B-1) ^ 2-4c)] / 2
x2=[-(b-1)+sqrt((b-1)^2-4c)]/2,
Because x1sqrt ((B-1) ^ 2-4c), we get (B-1) ^ 2 > (B-1) ^ 2-4c, so - 4c0
Re certification (2)
Because x2-x1

Calculating unknowns in mathematical problems
1 1 1
—— + —— = ——
x x+5 6
How much is x
It's a process
Online, etc
Well
Let's get to the original question
A section of road maintenance project on a certain highway is ready for external bidding. There are two engineering teams a and B bidding. The bidding data shows that: if the two teams cooperate, it can be completed in six days, and the total cost of the project is 10200 yuan; if the project is completed alone, team a takes 5 days less than team B, but the daily cost of team a is 300 yuan more than team B, The project headquarters decided to select one of the two teams to complete the project alone. If from the perspective of saving money, which team should be chosen? Why?
1/x+1/(x+5)=1/6

Note: 1 / 3 is used to represent one-third of X, and x ^ 2 is used to represent the square of X, so the original formula = 1 / x + 1 / (x + 5) = 1 / 6 general division. If both sides of the equation are multiplied by 6x (x + 5), then there is 6 (x + 5) + 6x = x (x + 5), so there is 6x + 30 + 6x = x ^ 2 + 5x shift term, x ^ 2-7x-30 = 0, cross method (X-10) (x + 3) = 0, so X-10 = 0 or x + 3 = 0, then x = 10 or x = -

lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+… +sin(nπ/n)]=?

Lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+… +Sin (n π / N)] observation: it can be seen that, in fact, the interval [0,1] is divided into N equal parts. For the function y = sin π x, the area is calculated at each interval point, and then the sum is obtained. Obviously, from the definition of definite integral, we can see that this and the definite integral ∫ sin π xdx from 0 to 1 are

The remainder of 555.5 divided by thirteen is

This number is equal to 5 (1 + 10 + 10 ^ 2 + 10 ^ 3 +...) +10^2008)=5(10^2008-1)/9
10 ^ 2008 = 1000 ^ 669 * 10, because 1001 = 7 * 11 * 13, the remainder of the formula is the same as (- 1) ^ 669 * 10, which is 3
So the remainder of 5 (10 ^ 2008-1) is 10
So the remainder of 5 (10 ^ 2008-1) / 9 is (13 * 2 + 10) / 9 = 4
I think this algorithm is more realistic. I made a mistake before. Sorry

-13 * 3 / 2-0.34 * 7 / 2 + 3 / 1 * (- 13) - 7 / 5 * 0.34 =? Simple calculation

-13*3/2-0.34*7/2+3/1*（-13）-7/5*0.34
=-13*3（1/2+1/1）-0.34*7（1/2+1/5）
=-39*3/2-0.34*7*7/10
=-58.5-1.666
=-60.166

Using undetermined coefficient method to find the general solution of differential equation y '' - 4Y '+ 4Y = (1 + X + x ^ 2 +... + x ^ 23) e ^ 2x (do not use differential operator),

Let the special solution of the equation be y * = P (x) e ^ λ x, then:
y''-4y'+4y=[p''+(2λ-4)p'+(λ^2 -4λ+4)p]e^λx =(1+x+x^2+...+x^23)e^2x
The characteristic equation has two roots: λ = 2; 2 λ - 4 = 0; λ ^ 2 - 4 λ + 4 = 0
P '' = 1 + X + x ^ 2 +... + x ^ 23 and: P (x) = x ^ 2 * q (23) (x) [Q (23) (x) is a polynomial of degree 23]
p(x)=x^3/2*3 + x^4/3*4 +...+x^25/24*25
General solution of differential equation:
y=e^2x(c1+c2*x + x^3/2*3+x^4/3*4+...+x^25/24*25)

Function domain y = LG (x ^ 2 + 3x-28)