a> B > 0, the minimum value of A2 + 1 / AB + 1 / a (a-b) is? Square of a

a> B > 0, the minimum value of A2 + 1 / AB + 1 / a (a-b) is? Square of a

The original formula = {(A & sup2; - AB) + [1 / (A & sup2; - AB)]} + {AB + [1 / (AB)]} ≥ 2 + 2 = 4. The equal sign is only obtained when a = √ 2, B = √ 2 / 2, so the minimum value of the original formula is 4

If a > 0, b > 0, and a + B = 1, then the minimum value of AB + 1ab is ()
A. 92B. 174C. 94D. 2

∵ a ＞ 0, B ＞ 0 by using the basic inequality, we can get that 1 = a + B ≥ 2Ab ∵ ab ≤ 14, let t = AB, then t ∈ (0,14] and y = t + 1t monotonically decrease in (0,14] ∵ when t = 14, the function has a minimum value of 174, so we choose B

Given a + b > 0, the process of a ` 3 + B ` 3 ≥ a ` 2B + AB ` 2 is proved

Because (a-b) ^ 2 ≥ 0,
That is, a ^ 2-AB + B ^ 2 ≥ ab
And a + B ≥ 0,
So (a + b) (a ^ 2-AB + B ^ 2) ≥ AB (a + b)
And a ^ 3 + B ^ 3 = (a + b) (a ^ 2-AB + B ^ 2)
Therefore, a ^ 3 + B ^ 3 ≥ a ^ 2B + AB ^ 2

PEP physics high school compulsory 2 Chapter 6 "gravity and space" all the formulas please
I'd like to make a special explanation for the knowledge related to quality,

Gravitation is a kind of interaction between objects due to the mass of objects. Its size is related to the mass of objects and the distance between two objects. The greater the mass of objects, the greater the gravitation between them; the farther the distance between objects, the smaller the gravitation between them, The following formula can be used to calculate: F = GMM / R ^ 2, that is, the universal gravitation is equal to the product of the gravitational constant times the mass of two objects divided by the square of their distance. Where g represents the gravitational constant, and its value is about 6.67 × 10 to the negative 11th power unit n · m2 / kg2, According to Kepler's second law, the angular velocity of planetary motion is certain, that is, ω = 2 π / T (period). If the mass of the planet is m, the distance from the sun is r, and the period is t, then according to the motion equation, the magnitude of the force on the planet is Mr ω ^ 2 = MR (4 π ^ 2) / T ^ 2, According to Kepler's third law, R ^ 3 / T ^ 2 = constant K ', then the force along the direction of the sun is Mr (4 π ^ 2) / T ^ 2 = MK' (4 π ^ 2) / R ^ 2. According to the relationship between the force and the reaction force, the sun is also subject to the force of the same magnitude as above. From the sun's point of view, (mass of the sun m) (k '') (4 π ^ 2) / R ^ 2 is the force along the direction of the planet, From the comparison of these two formulas, we can see that K 'contains the mass of the sun, m, and K' 'contains the mass of the planet, M. therefore, these two forces are proportional to the product of the mass of the two celestial bodies, which is called gravitation. If we introduce a new constant (called gravitation constant), and then consider the mass of the sun and the planet, as well as the previously obtained 4 · π 2, So it can be expressed as gravitation = (GMM) / (R ^ 2) the gravitation between two ordinary objects is so small that we can't notice it, so we can ignore it. For example, two people with a mass of 60 kg, 0.5 meters apart, the gravitation between them is less than one millionth of a Newton, and the force of an ant dragging a small straw is 1000 times that of the gravitation! However, in the celestial system, Due to the great mass of celestial bodies, universal gravitation plays a decisive role. The earth, which has a small mass in celestial bodies, has a great influence on the universal gravitation of other objects. It binds human beings, atmosphere and all ground objects to the earth, and makes the moon and man-made earth satellites rotate around the earth without leaving, The universal gravitation can be regarded as gravity, Mg = (GMM) / (R ^ 2), GM = g (R ^ 2), which is the golden substitution formula. And Mr ω ^ 2 = MR (4 π ^ 2) / T ^ 2 = mg

High school physics compulsory 2 Chapter 6 gravity and aerospace knowledge summary!
It's better to summarize each chapter in the book with clear thinking (∩)_ Thank you! No reference books

Commonly used
GMm/r^2=mr(2π/t)^2=(mv^2)/r=(mv2π)/T
=mrw^2
Density = 3G / 4 π RG (R is the radius of the planet)
mg=GMm/r^2
Application variant
Seeking the mass of celestial bodies (taking the calculation of the mass of the earth as an example)
① Know the period T, radius r of the moon around the earth
From GMM / R ^ 2 = MR (2 π / T) ^ 2
So, M = 4 (π ^ 2) (R ^ 3) / GT ^ 2
② Know the linear velocity V and radius r of the moon around the earth
From GMM / R ^ 2 = (MV ^ 2) / R,
So m = (RV ^ 2) / g
③ Know the speed limit V and period T of the moon around the earth
From GMM / R ^ 2 = (MV2 π) / T
So m = (2 π VR ^ 2) / TG = (TV ^ 3) / 2 π G
④ Know the radius r of the earth and the acceleration g of gravity on the earth's surface
Knowing M = GR ^ 2 / G from gold substitution (mg = GMM / R ^ 2)
Do the subject of gravitation, that is, simple celestial mechanics
Remember the formula is the most basic, many problems are set of formula
It's simple
To get high marks, look at the following
Here's what you need to pay attention to
1、 Establish two models
Determining the physical model of the research object is the first step to solve the problem, and the application of the law of universal gravitation is no exception. No matter how large the natural celestial bodies (such as the moon, earth, sun) or man-made celestial bodies (such as spacecraft, satellite, space station) are, first of all, they should be abstracted as particle models, In this way, the motion between them is abstracted as a uniform circular motion of one particle around another
2、 Grasp two ideas
There is a direct causal relationship between the gravity of objects and the motion of celestial bodies. Therefore, the law of gravitation is widely used in these problems
Idea 1: use the relationship between gravity and gravity
Namely
Train of thought 2: use the relationship between gravity and centripetal force
Namely
Where a is the centripetal acceleration, which can be expressed according to the conditions of the problem
3、 Distinguish three pairs of concepts
1. Gravity and universal gravitation
Gravity is produced by the attraction of the earth, but it is a component of universal gravitation. On the surface of the earth, it increases with the increase of latitude. Because there is little difference between the gravity of an object and the universal gravitation of the earth, it is generally considered that the two are equal in size. That is to say, at this time, this formula is called golden substitution, It can be expressed by the radius of the central star and its surface gravity acceleration
2. The centripetal acceleration with the rotation of the earth and the centripetal acceleration around the earth
The centripetal force of an object on the ground rotating with the earth is provided by the resultant force of the earth's gravity on the object and the supporting force on the ground; while the centripetal force of a satellite orbiting the earth is completely provided by the earth's gravity on it, and the values of the two centripetal forces are quite different, T is the rotation period of the earth; the centripetal acceleration of the satellite around the earth, where m is the mass of the earth and R is the distance between the satellite and the earth's center

The formula of universal gravitation in Physics

The formula of universal gravitation: 1. Kepler's third law: T2 / R3 = K (= 4 π 2 / GM) {R: orbital radius, t: period, K: constant (independent of the mass of the planet, depending on the mass of the central celestial body)} 2. The law of universal gravitation: F = gm1m2 / R2 (g = 6.67 × 10-11n & # 8226; m2 / kg2, direction on their line) 3

Physical universal gravitation formula
Please sum up the formula of gravity. It's better to type out what each letter represents - especially R, which formula represents the orbit radius and which formula represents the planet radius

Formula: F = gm1m2 / R ^ 2
G: Gravitational constant; M1, M2: the mass of two objects; R: the distance between two objects
When calculating the motion of planets around the sun and satellites around the earth, R represents the orbit radius;
When we calculate the first cosmic velocity and the density of the planet, R is equal to the radius of the planet
The key lies in the relationship between the two research objects

The physical formula of gravitation
Including deformation and so on,

Kepler's third law: T2 / R3 = K (= 4 π 2 / GM) {R: orbital radius, t: period, K: constant (independent of the mass of the planet, depending on the mass of the central celestial body)} 2. Law of gravitation: F = gm1m2 / R2, G = 6.67 × 10-11n? M2 / kg2, direction in their

Formula 2 of compulsory physics in Senior High School

(quoted from Ze Wuling)
It should be very detailed

Gauss's story, 350 words

The story of Gauss as a child
More than two hundred years ago, a 9-year-old's mathematical genius surprised his teacher
In 1787, in the third grade class of a rural primary school in Germany, a math teacher worked out a calculation problem:
　　1＋2＋3＋4＋5＋… ＋98＋99＋100.
It's a test for the third grade students to add up 100 numbers one by one
Unexpectedly, as soon as the teacher finished the topic, one of the students in the class, named Gauss, handed in his slate
At first, the teacher didn't care. He handed it in so soon. Who knows what he wrote?
Later found that the class only one person to do right, that is, the fast turnover of Gauss
Gauss's solution surprised the teacher even more
Gauss took one of these 100 numbers from both ends to the middle, and matched them with 1 and 100, 2 and 99, 3 and 98 The sum of two numbers in each pair is equal to 101, so the original formula is 101 × 50 = 5050
Although this algorithm was not initiated by little Gauss, no one had taught him in advance. In Germany more than two hundred years ago, this calculation method was taught in universities, which is called summation of arithmetic progression. Even today, with the rapid development of science and technology, summation of arithmetic progression progression has to be systematically studied in mathematics class of senior high school. Gauss, who was only 9 years old, was born in a poor family, The teacher's name is pumitenell. He went to the big city Hamburg to buy a mathematics book for Gauss to read, and asked his young assistant Barthes to take care of Gauss
Later, Gauss continued to study hard and made many important contributions in mathematics, astronomy and physics. He was known as the "king of mathematicians" and was as famous as Archimedes and Newton. Gauss was an eternal star in the history of mathematics