1 + 2 + 3 +,,, + 2012 + 2013 Simple, clear. Thank you

1 + 2 + 3 +,,, + 2012 + 2013 Simple, clear. Thank you

This formula has a rule, 1 + 2013 = 2 + 2012 = 3 + 2011. = 1006 + 1008 = 2014. There are 1006 groups in total, and so on
And then there's a 1007 in the middle of the equation, so
1+2+3+、、、+2012+2013=
1006 times 2014 + 1007 = 202407

How about 1 + (- 2) + 3 + (- 4) + 5 +... + (- 2012) + 2013

For example, add (- 2) and 1 to get (- 1), and then add the following two numbers to get (- 1), so the last negative number is (- 2012). Then divide it by two to get 1006, which is the total number of negative one after all pairwise addition. All negative ones add up to (- 1006), and then add (- 1006) to 2013
1007······

(-0.125）^2013*8^2014+(-1)^2012+(-1)^2011

(-0.125）^2013×8^2014+(-1)^2012+(-1)^2011
＝﹙﹣1／8﹚^2013×8^2013×8＋1＋﹙﹣1﹚
＝﹛﹙﹣1／8﹚×8﹜^2013×8
＝﹙﹣1﹚×8
＝﹣8

Index finger, ring finger, thumb, little finger, Chinese finger?

Index finger, ring finger, thumb, pinky, middle finger
There is no the in front of the English translation

If x and Y belong to R, 3x ^ 2 + 2Y ^ 2 = 6, find the minimum value of X + Y -- the maximum value of x ^ 2 + y ^ 2
Wrong. It should be the maximum value of X + y and the minimum value of x ^ 2 + y ^ 2

The maximum value of X + y is √ 5, and the minimum value of x ^ 2 + y ^ 2 is 2
3x＾2＋2y＾2＝6
y^2=(6-3x^2)/2
t=x^2+y^2
=x^2+(6-3x^2)/2
=x^2+3-3x^2/2
=-x^2/2+3.
It is easy to know that when x = 0, t has a maximum value of T = 3
When 3x ^ 2 = 6, that is, x ^ 2 = 2, there is a minimum value, t = - 2 / 2 + 3 = - 1 + 3 = 2
Let x + y = t
y=t-x
So 3x ^ 2 + 2 (t-x) ^ 2 = 6
5x^2-4tx+2t^2-6=0
△=16t^2-20(2t^2-6)≥0
4t^2-5(2t^2-6)≥0
6t^2≤30
t^2≤5
-√5≤t≤√5

1.3.9.19 The general formula of the composition/

The general formula: an = 2 times n square minus 4 times n plus 3

It is known that x = 3, y = 1 and x = 1, y = - 3 are solutions of binary linear equations ax + by = 5 with respect to XY. When x = 2, y = 5, find the value of ab

1) First, the two pairs of values are substituted into the system of equations, so it is a system of linear equations with two variables, and the solution is a = 2, B = - 3
2) Then put a = 2, B = - 3 from the above problem into the formula, and then put x = 5, y = - 1 in the problem, and the value is 13

Four students went to plant trees. The first student planted half of the total trees planted by other students. The second student planted 13 of the total trees planted by other students. The third student planted 14 of the total trees planted by other students. Then the fourth student just planted 13 trees. How many trees did the four students plant?

The four students have planted 60 trees in total

Let f (x) = LX + 1L + LX + 2L + +lx+2011l+lx-1l+lx-2l+…… +Lx-2011l (x ∈ R), if x0 is the equation f (x ^ 2-3x + 2) = f (x-1)
Then the sum of all integers x0 satisfying the condition is______ .

Suppose X1 = x ^ 2-3x + 2, X2 = X-1
In short, we can deduce from the original formula that if we want the equation to hold, then the difference between | x1 | and | x2 | must be less than 1, that is, 0
1.x1+x2=0
X ^ 2-3x + 2 + X-1 = x ^ 2-2x + 1, x = - 1 and 1. Obviously, x = - 1 needs to be discarded
2.x1-x2=0
X ^ 2-3x + 2-x + 1 = x ^ 2-4x + 3 = (x-3) (x-1), x = 3
3. In another special case, the difference between | x1 | and | x2 | is 1, and the equation still holds, provided that one is 1 and the other is 0
X2 = 0, then x0 = 1, X1 = 0, impossible
X1 = 0, x0 = 1 or 2,1 has been discussed, discard, x0 = 2, then x2 = 1, the equation holds
So, 1 + 3 + 2 = 6

The capacity of each bottle of a beverage is 750 mL, and each bottle is 24 yuan,
1) Auntie Wang is going to buy 20 bottles. How much is the actual cost? (2) how many ml are the 20 bottles in total? How many liters are they?

1) In fact, it only costs 96 yuan. Because you can get one free if you buy four bottles and four free if you buy 16 bottles, you can use 24 * 4 + = 96
2) 20 * 750 = 15000 ml = 15 L