If we know that M minus n equals 1 / 5, then how much is one-third multiplied by bracket n minus bracket m

If we know that M minus n equals 1 / 5, then how much is one-third multiplied by bracket n minus bracket m

Given M-N = 1 / 5, find 1 / 3 (n-m) =?
From the problem, N-M = - (m-n) = - 1 / 5
1/3（n-m）= （1/3）X（-1/5）= -1/15
PS: in other words, what grade is it? It's so simple. LZ is wandering in class!

In the triangle ABC, a ^ 2 + C ^ 2-B ^ 2 = √ 3aC, find the angle B

a^2+c^2-b^2=√3ac
cosB=(a^2+c^2-b^2)/2ac=√3/2
So B = 30 degrees

In machine number (), the representation of zero is unique
A original code B complement code c shift code D inverse code

B
The complement is the inverse plus one
Such as 100000000
Inverse code 11111111
Complement 11111111 + 1 = 100000000

1+2-3+4-5+6-7+8-9…… +2012-2013=?

-1006

Let f (x) have f derivative (x) in the open interval (a, b)

F '(x) 0 indicates that the function is concave
So the answer is C

(1 / 1 * 2 + 1 / 2 * 3 + 1 / 3 * 4 +. + 1 / 2004 * 2005)

1/n*(n+1)= 1/n - 1/(n+1)
So the above formula is equal to 1-1 / 2 + 1 / 2-1 / 3 +... + 1 / 2004-1 / 2005 = 1-1 / 2005 = 2004 / 2005

If the point P1 (2, - 3) is shifted to the right by 3 unit lengths, and then down by 2 unit lengths to the point P2, then the coordinate of P2 is ()
A. （5，-1）B. （-1，-5）C. （5，-5）D. （-1，-1）

∵ point P1 (2, - 3) moves 3 units of length to the right, and then moves 2 units of length down to point P2. The abscissa of P2 is 2 + 3 = 5, and the ordinate is - 3-2 = - 5, so select C

1; 1 / 2 x + 3 / 1 x = 4 / 3 2; (1-25%) x = 36
1. 1 / 2 x + 1 / 3 x = 3 / 4
2;（1-25%）x=36

(3X+2X)/6=3/4
5X/6=3/4
20X=18
X=0.9
0.75X=36
3/4X=36
X=48

Let A1, A2, A3, A4 and A5 be five different points on the plane, then the number of points m with MA1 + ma2 + Ma3 + ma4 + Ma5 = 0 is ()
A. 0B. 1C. 5D. 10

According to the meaning of the question, let the coordinates of M be (x, y), and the number of groups be obtained by solving x, y, that is, the number of points m that meet the conditions, and then let the coordinates of A1, A2, A3, A4, A5 be (x1, Y1), (X2, Y2), (X3, Y3), (x4, Y4), (x5, Y5) in turn; if MA1 + ma2 + Ma3 + ma4 + Ma5 = 0, then (x1-x, y1-y) + (x2-x, y2-y) + (x3-x, y3-y) + (x4-x, y4-y) + (x5-x, y5-y) ）=0, then x = X1 + x2 + X3 + X4 + x55, y = Y1 + Y2 + Y3 + Y4 + y55; there is only one set of solutions, that is, there is only one and only one qualified point m; so B

1、 Factorization factor 1. X ^ 2 (X-Y) + (Y-X) 2. (x + y) ^ 2-14 (x + y) + 493.7x ^ n + 1-14x ^ n + 7x ^ n-1 (n is an integer not less than 1)
It can't finish 4. M ^ 2 + n ^ 2 + 4m-6n + 15 5. (X-2) ^ 2 + 10 (X-2) + 25 2. Try to prove that no matter m.n takes any number, the value of the algebraic formula m ^ 2 + n ^ 2 + 4m-6n + 15 is always positive number 3. Calculate 2 ^ 2012-2 ^ 2011-2 ^ 2010 4. Given that 1 / 2 of x-x = 2, find 1. X ^ 2 + 1 / 2 of x ^ 4 2. X ^ 4 + 1 / 3. X ^ 8 + 1 / 8 of x ^ 8
To write the process, I use it in the evening

1. x^2(x-y)+(y-x)=（x-y）（x-1）（x+1）
2.（x+y)^2-14(x+y)+49 =(x+y-7）2
3. 7x ^ n + 1-14x ^ n + 7x ^ n-1 (n is an integer not less than 1) = 7xn-1 (x-1) 2
4.m^2+n^2+4m-6n+15 =(m+2)2+(n-3)2
5.(x-2)^2+10(x-2)+25=（x+3)2
2、 Try to prove that the value of the algebraic formula m ^ 2 + n ^ 2 + 4m-6n + 15 is always positive no matter what number m.n takes
=(m+2)2+(n-3)2
(M + 2) 2 is greater than or equal to 0, (n-3) 2 is greater than or equal to 0
3、 Calculation 2 ^ 2012-2 ^ 2011-2 ^ 2010
=22010（22-2-1）=22010
4、 Given 1 / 2 of x-x = 2, find 1. X ^ 2 + 1 / 2 of x ^ 2. X ^ 4 + 1 / 4 of x ^ 3. X ^ 8 + 1 / 8 of x ^ 8 (x-1 / x)
If 1 / 2 of x-x = 2 is squared, then x ^ 2 + 1 / 2 of x ^ 2 equals 6
If x ^ 2 + 1 / 2 of x ^ 2 = 6, the square x ^ 4 + 1 / 4 of x ^ 2 is equal to 34
If x ^ 4 + x ^ 4 / 1 = 34, the square ^ 8 + x ^ 8 / 1 equals 1154
22010 is the 2010 power of 2, and there are many 2's that are quadratic