The differential of y = 1 / √ LNX

The differential of y = 1 / √ LNX

y=(lnx)^(-1/2)
y'=-1/2*(lnx)^(-3/2)* (lnx)'=-1/(2x)*(lnx)^(-3/2)
So dy = - DX / (2x) * (LNX) ^ (- 3 / 2)

As shown in the figure, the parabola y = - x2 + 2 (M + 1) x + m + 3 intersects the x-axis at two points a and B. If OA: OB = 3:1, find the value of M______ ．

Let B (- K, 0), then a (3k, 0); - K, 3K are two of the equations - x2 + 2 (M + 1) x + m + 3 = 0,; − K + 3K = 2 (M + 1) − K · 3K = − (M + 3). The solution is: M = 0 or - 53, ∵ all satisfy △ 0, as shown in the figure: if X1 and X2 are two of the equations - x2 + 2 (M + 1) x + m + 3 = 0, then x1 · x2 = - (M + 3) < 0, X1 + x2 = 2 (M + 1) > 0, when m = - 53, X1 + x2 = 2 (M + 1) = - 43 < 0, ∵ M = - 53 is not proper In this paper, we present a new method to solve the problem

-(- 1) power 2012 - 1 / 36 × [| - 2's third power | - 12 + (- 2's third power]

-(-1)^2012-1/36×[▏-2▏-12+(-2)³]=-1-1/36×[2-12-8]=-1-1/36×(-18)
=-1+1/2=-1/2

Problems of polar coordinate equation
In polar coordinate system, the polar coordinate equation of circle C is p = 2sina. The polar coordinate equation of a line passing through a point P (2, π / 2) on the circle and tangent to the circle is solved. A line L passing through a few points intersects the circle at O and a, and the angle AOX = 60 ° to find the length of OA [add points to the process]
Wrong type. A straight line L passing through the "pole" intersects the circle at O and a, and the angle AOX = 60 ° to find the length of OA [add points to the process]

ρ^2=2ρsinA
x^2+y^2=2y
x^2+(y-1)^2=1
The tangent through the point is y = 2
The polar equation is ρ sin Θ = 2
According to the rectangular coordinates of circle C, the center of circle is (0, - 1)
Because the circle C is tangent to the X axis, the central angle of OA is 120 degrees
∠ COA = ∠ Cao = 30 degrees
R=1
So OA = 2 √ 3

2 + 2 × 2 × 3 + 3 & # 178; factorization by formula method
I don't feel like following the routine

Factorization is relative to polynomials. You are a number, which has nothing to do with factorization. If you have to decompose according to factorization, you can take the following steps
2 + 2 × 2 × 3 + 3 & # 178; + 2 - 2 = (3 + 2) &# 178; - 2 = (3 + 2-radical 2) (3 + 2 + radical 2)

If the points (1, Y1) and (- 2, Y2) are on the function y = - 2x + 1 image, the relationship between Y1 and Y2

According to the question:
-2＋1＝y1
y1＝-1
4＋1＝y2
y2＝5
Then: Y2 > Y1
　　y1＝y2－6
Don't understand can ask!

To solve the equation: 1 / 3 X-2 and 2 / 3 = 1 / 4 x + 3 x are unknowns, not multipliers

X is 32

It is known that a (- 2,2) is a point on the positive proportional function y = KX
Ask if there is a point P on the x-axis, so that A.P and the origin of coordinates o form an isosceles right triangle? If there is, try to find the coordinates of point p; if not, please explain the reason. (please finish this problem, explain how to do this kind of problem, it's better to find some of these types of problems ~)

A1 is (2,0), A2 is (4,0),
To solve this kind of problem, first of all, you need to read the title and understand what the title is for and what the requirements are
After defining the conditions, draw the coordinate axis. According to the conditions, you can know that y = - x, through two or four quadrants, forms a 45 degree angle with the X axis. Mark point P on the straight line, determine that there is an angle of 45 degrees, and then find the angle of 90 degrees. At this time, the classification is divided into OP as the hypotenuse and X axis as the hypotenuse, draw the angle of 90 degrees, and then extend the straight line to find the coordinates of the intersection point with the X axis
PS: don't blame me for some unclear points. I'm a student, too. Although I'm much older than you

There are several ways to solve binary linear equations

Three?
Substitution method
Integral approach
cross multiplication
That's all I know -- 0

The calculation method of Ohm's law of the whole circuit, please

Ohm's law of whole circuit (Ohm's law of closed circuit) I = E / (R + R) where e is electromotive force, R is internal resistance of power supply, internal voltage uine = IR, e = uine + uexternal scope of application: energy conversion in closed circuit of pure resistance circuit: e = U + IR EI = UI + I ^ 2R P release = EI P output = UI, P output in pure resistance circuit = I ^ 2R = e ^ 2R / (...)