[calculus] let z = Z (x, y) satisfy the system of equations f (x, y, Z, t) = 0, G (x, y, Z, t) = 0, where f and g have continuous partial derivatives

The partial derivative of F with respect to the i-th variable is expressed by FI, I = 1,2,3,4df (x, y, Z, t) = 0, that is, f1dx + f2dy + f3dz + f4dt = 0, ① DG (x, y, Z, t) = 0, that is, g1dx + g2dy + g3dz + g4dt = 0, ② × G4 - ② × F4, then (f1g4-g1f4) DX + (f2g4-g2f4) dy + (f3g4-g3f4) DZ = 0, so DZ = - [(f1g4-g1f4) d

X ^ 2 + y ^ 2 + Z ^ 2 = 2Z, φ (x, y) = 0 and continuous partial derivative, find DZ / DX

Let f = 2Z-X ^ 2-y ^ 2-z ^ 2
Fz=2-2z
Fx=-2x
az/ax=-Fx/Fz=x/（1-z）

If AB = 5cm, BC = 20cm, AC = 15cm, then points a, B, C are on the same line
Is that right? Why

If the premise is the same plane, it is true
The order of the points is B, a, C
According to the triangle theorem, the sum of any two sides is greater than the third side
And ab + BC = AC in the question, so the three points can't form a triangle
In a plane, if three points cannot form a triangle, the three points must be collinear, that is, on a straight line

In the reactive power calculation formula, what is the sin value? In the compensation capacity calculation formula, q = P * (tg1-tg2), TG = sin / cos,
How to calculate the inverse function sin in reactive power compensation

Let a > 0, a ≠ 1 and f (x) = loga (x2-2x + 3) have the minimum value, then the solution set of inequality loga (x-1) > 0 is______ ．

From a ＞ 0, a ≠ 1, the function f (x) = loga (x2-2x + 3) has the minimum value, we can know that a ＞ 1, so the inequality loga (x-1) ＞ 0 can be transformed into X-1 ＞ 1, that is, X ＞ 2

In RT △ ABC, CD is the height of hypotenuse AB, and AC: BC = 3:4, then ad: BD=_________ .

ABC ∽ ADC ∽ BDC, AC / ad = BC / CD, AB / BC = AC / CD, similarly AC / CD = CD / BD
And AC / BC = ad / CD = CD / BD = 3 / 4, so ad: BD = 9:16

If the main argument value of complex number (1 + AI) ^ 2 is π / 4, then the value of real number a is π / 4

∵ (1 + AI) ^ 2 = 1 + 2ai-a ^ 2 = (1-A ^ 2) + 2ai, and the main value of (1 + AI) ^ 2 is π / 4,
Let 1-A ^ 2 = RCOs (π / 4) = R / √ 2, 2A = rsin (π / 4) = R / √ 2, where R is the module of complex number (1 + AI) ^ 2
1-A ^ 2 = 2A, a ^ 2 + 2A + 1 = 2, a (a + 1) ^ 2 = 2, a + 1 = √ 2, or a + 1 = - √ 2,
A = - 1 ＋ 2, or a = - 1 ＋ 2

Given that circle C1: X & # 178; + Y & # 178; - 2ax-2y + A & # 178; - 15 = 0, circle C2: X & # 178; + Y & # 178; - 4ax-2y + 4A & # 178; = 0 (a > 0), try to find the value of a when two circles C1 and C2
(1) There is a unique common point (2) there are two common points (3) there is no common point

C1: (x-a) &# 178; + Y & # 178; = 15 radius of center (a, 0) R1 = root 15
C2: (x-2a) & #178; + (Y-1) & #178; = 1 radius of center (2a, 1) R2 = 1
1) The only common point --- tangency --- R1 + R2 = | C1C2 | or R1-R2 = | C1C2 ||
2) Two common points -------- intersection -------- R1 + R2 greater than | C1C2|
3) There is no common point --- separation or separation of one circle in another circle: R1 + R2 is less than | C1C2 | I didn't think of a circle in another circle

Indefinite integral calculation ∫ DX / [(2x & # 178; + 1) √ (X & # 178; + 1)]
Is this calculation correct?
Let t = √ (X & # 178; + 1), then T & # 178; = x & # 178; + 1 DX = DT
The original formula = ∫ DT / {[2 (T & # 178; - 1) + 1] t}
=∫dt/[(2t²-1)t]
=∫[2t/(2t²-1)-1/t]dt
=∫[2t/(2t²-1)]dt-∫(1/t)dt
=(1/2)∫[1/(2t²-1)]d(2t²-1)-∫(1/t)dt
=(1/2)ln(2t²-1)-lnt+C
And bring x back
If x = constant, the results are quite different
Which step is wrong?

It was wrong from the beginning
T = √ (X & # 178; + 1), then T & # 178; = x & # 178; + 1
x=√t^2-1
dx/dt=t/(√t^2-1)

English translation
With the coming of World War II, many eyes in implied Europe turned hopefully, or desperately, toward the free American continuous. 2, I'm through with everything here. 3, the smiling young prince showed no sign of the strain of the week's continuous public appearances, There was a land of Cavaliers and cotton fields called the old South Here in this pretty world Gallantry took its last bow… Here was the last ever to be seen of knights and their ladies fair of master and of slave… Look for it only in books,for it is no more than a dream remembered… A civilization gone with the wind… This is the test content of our elective course,

With the coming of the Second World War, some prisoners in Europe showed hope and some despair when they went to the American continent. 2. I'm tired of everything here. 3. In the diplomacy of a week in a row, the little princess who always smiles didn't feel nervous and tired at all. 4. I believe in the relationship between people

[calculus] let z = Z (x, y) satisfy the system of equations f (x, y, Z, t) = 0, G (x, y, Z, t) = 0, where f and g have continuous partial derivatives