How to calculate 49999 + 4999 + 1999 + 499 + 49 + 5? How to calculate 382-98? Another one: (215 + 357 + 429 + 581 + 625) - 205 + 347 + 419 + 571 + 615) is also simple!

How to calculate 49999 + 4999 + 1999 + 499 + 49 + 5? How to calculate 382-98? Another one: (215 + 357 + 429 + 581 + 625) - 205 + 347 + 419 + 571 + 615) is also simple!

49999+4999+1999+499+49+5= (49999+1)+(4999+1)+(1999+1)+(499+1)+(49+1)= 50000 + 5000 + 2000 + 500 + 50= 57550382-98= 382 - 100 + 2= 282 + 2= 284 (215+357+429+581+625)-(205+347+419+571+615)= (215-205)+(3...

100 points for a math problem,
E ^ (x ^ 2) = e ^ (15x-50) find out the two values of X. what are the bigger and the smaller ones

x^2=15x-50
x^2-15x+50=0
(x-5)(x-10)=0
x=5,x=10

The distance between city a and city B is 55 kilometers. Wang Ming started from city a to city B, walked 25 kilometers first, and then changed to riding a bicycle. The speed doubled. When he arrived in city B, he found that the walking time was one hour longer than the cycling time. Unfortunately, Wang Ming's speed was () km / h?
There's another one,
For the integer 6 (x ^ 5) + 5 (x ^ 4) + 4 (x ^ 3) + 3 (x ^ 2) + 2x + 2002, given a numerical value, if Xiao Ming calculates the value of the integer according to the rules of four operations, he needs to calculate 15 times of multiplication and 5 times of addition, Xiao Hong said: there is another algorithm, as long as the appropriate brackets are added, the number of addition can be kept unchanged, while the multiplication only counts 5 times.
Is Xiaohong right or wrong? If it is right, please explain the reason: what does Xiaohong think

Walking 25 kilometers, the speed is x, riding 30 kilometers, the speed is 2x
25/x－1＝30/2x
The solution is x = 10
So the walking speed is 10 kilometers per hour
The second question is: {[(6x + 5) x + 4] x + 3} x + 2 ＇ x + 2002
Edit it

If f (1) > F (2) = (2a-3) / (a + 1), the value range of a is obtained

f(1)=-f(-1)=-f(-1+3)=-f(2)
So - f (2) > F (2)
So f (2)

If the function f (x) is odd and the period of F (3x + 1) is 3, f (1) = - 1, find f (2006)

If f (3x + 1) period is 3, then
F (3x + 1) = f (3 (x-3) + 1) = f (3x-8), so f (x) period is 9
F (2006) = f (9 * 223-1) = f (- 1), and f (x) is an odd function, f (- 1) = - f (1) = 1
So f (2006) = 1

If the function f (x) is odd and the period of F (3x + 1) is 3, f (1) = - 1, then f (2006) is equal to ()
A. 0B. 1C. -1D. 2

The period of ∵ f (3x + 1) is 3 ∵ f (3x + 1) = f [3 (x + 3) + 1] = f (3x + 1 + 9), that is, the period of F (T + 9) = f (T) ∵ function f (x) is 9 ∵ f (2006) = f (9 × 223-1) = f (- 1), and f (x) is an odd function, f (- 1) = - f (1) = 1, so select: B

It is known that the function y = f (x) defined on R is odd, y = f (x + 1) is even, f (1) = 1, and the period of F (x) is obtained

From F (x) = - f (- x), f (x + 1) = f (- x + 1)
Then f [(x + 1) + 1] = f [- (x + 1) + 1] = f (- x) = - f (x) = - f (x)
f(x)=f[(x-1)+1]=f[-(x-1)+1]=f(-x+2)=-f(x-2)
So - f (X-2) = - [- f (x)] = - f [(x + 1) + 1] = - f (x + 2)
So f (X-2) = f (x + 2)
Then f (x) = f (x + 4), so the period T = 4

Let f (x) be an odd function whose period is 5

If the period is 5, then f (x) = f (5N + x), n ∈ Z, so f (2013) = f (43 × 5-2) = f (- 2) because it is an odd function, f (- x) = - f (x), so f (- 2) = - f (2) when 0

Given that f (x) is an odd function defined on R, its minimum positive period is 3, tangent x ∈ (- 3 / 2,0), f (x) = log (1 / 2) (1-x), then f (2013) + F (2014)
Log (1 / 2) (1-x)

f(x+3n)=f(x) (n∈Z)
When x = - 1, f (x) = Log1 / 2 (1 - (- 1)) = Log1 / 2 (2) = Log1 / 2 (1 / 2) ^ (- 1) = - 1
When x = 0, f (x0 = Log1 / 2 (1-0) = 0
f(1)=-f(-1)=1
f(2)=f(-1+3)=f(-1)=-1
f(3)=f(0+3)=f(0)=0
f(2013)=f(0+3*671)=f(0)=0
f(2014)=f(1+3*671)=f(1)=1
f(2013)+f(2014)=0+1=1

2. Let f (x) (x ∈ R) be an odd function with period 3, and f (1) > 1, f (2) = a, then () (a) a > 2; (b) a < - 2; (c) a > 1; (d) a < - 1

Because f (2) = a, f (x) is a function with period 3, so f (2) = f (2-3) = f (- 1) = a
Because f (x) is an odd function, f (1) = - f (- 1) = - a > 1
So a