Given 14 (x 2Y 3) m (2xyn-3) 2 = x 4Y, find the value of (M 2n) 3

∵ 14 (x2y3) m (2xyn-3) 2 = 14 (x2my3m) (4x2y2n-6) = x2m + 2y3m + 2n-6 = x4y, ∵ 2m + 2 = 43M + 2n − 6 = 1, M = 1n = 2, ∵ m2N) 3 = 8

（1）-a3•a4•a+（a2）4+（-a4）2 （2）（-3x2y）2 （-2xy3z）3（3）（5a2b-3ab-1）（-3a2）（4）6x2-（x-1）（x+2）-2（x-1）（x+3）（5）5m×125m÷252m-1 （6）[x+2y）（x-2y）+4（x-y）2]÷6x．

（1）-a3•a4•a+（a2）4+（-a4）2 =-a8+a8+a8 =a8；（2）（-3x2y）2 （-2xy3z）3=9x4y2 （-8x3y9z3）=-72x7y11z3；（3）（5a2b-3ab-1）（-3a2）=-15a4b+9a3b+3a2；（4）6x2-（x-1）（x+2）-2（x-1）（x+3）=6x2-（x2+x-2）-2（x2+2x-3）=6x2-x2-x+2-2x2-4x+6=3x2-5x+8；（5）5m×125m÷252m-1 =5m×53m÷54m-2=25；（6）[x+2y）（x-2y）+4（x-y）2]÷6x=[x+2y）（x-2y）+4（x-y）2]÷6x=[x2-4y2+4x2-8xy+4y2]÷6x=[5x2-8xy]÷6x=56x-43y．

（1）-a3•a4•a+（a2）4+（-a4）2 （2）（-3x2y）2 （-2xy3z）3（3）（5a2b-3ab-1）（-3a2）（4）6x2-（x-1）（x+2）-2（x-1）（x+3）（5）5m×125m÷252m-1 （6）[x+2y）（x-2y）+4（x-y）2]÷6x．

（1）-a3•a4•a+（a2）4+（-a4）2 =-a8+a8+a8 =a8；（2）（-3x2y）2...

Calculation. (1) (a-2b + 3C) 2 - (a + 2b-3c) 2; (2) [AB (3 − b) − 2A (B − 12b2)] (− 3a2b3); (3) - 2100 × 0.5100 × (- 1) 2013 △ (- 1) - 5; (4) [(x + 2Y) (x-2y) + 4 (X-Y) 2-6x] △ 6x; (5) 5A2 - [A2 + (5a2-2a) - 2 (a2-3a)]

（1）（a-2b+3c）2-（a+2b-3c）2=（a-2b+3c+a+2b-3c）（a-2b+3c-a-2b+3c）=2a•（-4b+6c）=12ac-8ab；（2）[ab(3−b)−2a(b−12b2)](−3a2b3)=[3ab-ab2-2ab+ab2]（-3a2b3）=ab（-3a2b3）=-3a3b4；（3）-2100×0.5100×（-1）2013÷（-1）-5=-（2×0.5）100×（-1）÷（-1）=-1×（-1）÷（-1）=-1；（4）[（x+2y）（x-2y ）+4（x-y）2-6x]÷6x=[（x2-4y2）+4（x2-2xy+y2）-6x]÷6x=[x2-4y2+4x2-8xy+4y2-6x]÷6x=[5x2-8xy-6x]÷6x=56x-43y-1；（5）5a2-[a2+（5a2-2a）-2（a2-3a）]=5a2-[a2+5a2-2a-2a2+6a]=5a2-[4a2+4a]=a2-4a．

Ysinx + cos (X-Y) = 0, find dy / DX | (x = π / 2)

On both sides, take the derivative of X: dy / dxsinx + ycosx sin (X-Y) (1-dy / DX) = 0, take x = π / 2 into the known equation to get y, and then take X and Y into the above equation to get the result

The application of plane vector
In the isosceles triangle ABC, BD is the midline on the AC waist (D is the midpoint of AC), CE is the midline on the AB waist (E is the midpoint of AB), and BD is perpendicular to CE, so the cosine value of vertex angle a can be calculated?
I haven't learned cosine theorem yet!
Say to use vector to solve problems

Let the intersection of BD and CE be f, then f is the center of gravity
If AF is connected and BC is extended at point G, then G is the midpoint of BC, Ag = 2gf (theorem of center of gravity)
Because of isosceles, Ag is perpendicular to BC, so the triangle FGC is isosceles right triangle
The triangle AGC is a right triangle. You can find the BC side,
With AB, AC, BC, trilateral, cosa can be obtained by cosine theorem

Function: given f (x) = cos2x + asinx + 3, find the range of F (x)

f(x)=cos²x-sin²x+asinx+3=-2sin²x+asinx+4=-2(sinx-a)²+4+a²
So f (x)

Primitive function with derivative 2 * x + X * 3

Is that power?
∫(2*x+x*3)dx
=∫2*xdx+∫x*3dx
=1/ln2∫2*xdx+x*4/4
=2*x/ln2+x*4/4+C

Xiao Ming and Xiao Gang are good friends. They go to a grain and oil store to buy oil twice in a month. The price of oil has changed twice. The first one is x yuan / kg, and the second one is y yuan / kg. Their buying ways are different: Xiao Ming always buys the same amount of oil each time, while Xiao Gang only takes out the same amount of money to buy oil each time?

Suppose Xiao Ming bought 2A kg of oil twice and 4A kg of oil altogether;
The total cost is 2aX + 2ay = 2A (x + y)
If Xiaogang also spent 2A (x + y) yuan in two times, and spent a (x + y) yuan in two times respectively,
A (x + y) / x + a (x + y) / y = a (x + y) & sup 2; / (XY) kg of oil can be bought
Because, from (X-Y) & sup2; ≥ 0, we can get: (x + y) & sup2; ≥ 4xy,
Therefore, (x + y) & sup2; / (XY) ≥ 4, we can get a (x + y) & sup2; / (XY) ≥ 4a
It also costs 2A (x + y) yuan,
When x ≠ y, Xiao Gang bought more oil than Xiao Ming,
When x = y, two people buy the same amount of oil;
Therefore, Xiaogang's way of buying oil is cost-effective

Given 14 (x 2Y 3) m (2xyn-3) 2 = x 4Y, find the value of (M 2n) 3