The story of soldiers on paper

The story of soldiers on paper

According to the biography of Lin Xiangru in historical records, Zhao Kuo, the son of Zhao she, a famous general of Zhao state in the Warring States period, learned the art of war when he was young, and his father could not defeat him when he talked about military affairs. Later, he took over from Lian Po as Zhao general, and in the battle of Changping, he only knew how to do things according to the military books, but did not know how to change the way, and was defeated by the Qin army

A story about soldiers on paper
Don't talk about the story on paper, but some other stories, which are similar to the story on paper. The main idea is to understand the theory without practice and then fail

The most typical is "ZHUGE Liang cut Ma Su in tears", Ma Su's paper, lack of practical experience

Seeking the story of "war on paper"
Just a brief summary, not too much detail

During the Warring States period, Zhao Kuo, the son of Zhao she, a famous general of the Warring States period, was full of military knowledge. He was able to talk about how to use military. Even his father thought he was invincible. Zhao she thought he was a soldier on paper and didn't know how to change his mind. Later, Zhao she died and Zhao Kuo took the place of Lian Po. Lin Xiangru and others tried their best to fight against him. The king of Zhao insisted that Zhao Kuo lost 400000 soldiers in the battle of Changping

How to analyze the orthogonal decomposition of force after drawing

The first step is to select the research object and express it in the form of particle. The second step is to analyze the force on the selected research object. The third step is to establish the rectangular coordinate system. Generally speaking, the coordinate system can be established arbitrarily in the horizontal plane, but it is better to build along the sliding direction of the object on the inclined plane

6（p+q)^-2(p+q)
2(x-y)^-x(x-y)
m(a-b)-n(b-a)
121*0.13+12.1*0.9-1.2*12.1
2.34*13.2+0.66*13.2-26.4
Sorry, it's the second power

(1) Original formula = 2 (P + Q) (3P + 3q-1)
(2) Original formula = (X-Y) (2x-2y-x) = (X-Y) (x-2y)
(3) The original formula = (a-b) (M + n)
(4) Original formula = 12.1 × (1.3 + 0.9-1.2) = 12.1
(5) Original formula = 13.2 × (2.34 + 0.66-2) = 13.2
These questions should be very simple, study hard and make progress every day
I wish you success in your studies*^__ ^*)

(1).x^2-8xy+7y^2
(2).4x^2+9xy+2y^2
(3).2x^2+7xy+3y^2
(4).6x^2+xy-2y^2

(1).x^2-8xy+7y^2 =（x-y)(x-7y)
(2).4x^2+9xy+2y^2 =(4x+y)(x+2y)
(3).2x^2+7xy+3y^2 =(2x+y)(x+3y)
(4).6x^2+xy-2y^2=(2x-y)(3x+2y)

A factoring problem
1. Quadratic power of Y (X-Y) - (cubic power of Y-X)

Because the quadratic power of (X-Y) = the quadratic power of (Y-X)
So, the formula can become
The second power of Y (Y-X) minus the third power of (Y-X)
Extracting the quadratic power of the common factor (Y-X)
What we got was
The second power of (Y-X) multiplied by [y - (Y-X)]
The second power of X (Y-X) is obtained

Answer in 10 minutes and add 20 points!
It is known that a, B and C are three sides of the triangle ABC
A ^ 2 + B ^ 2 + C ^ 2-AB BC AC = 0 (a twice + B twice + C twice - AB BC AC equals 0)
Judge the shape of triangle ABC and explain the reason
The reason is the most important!
This is the second way:
When a is a, the time-sharing equation 3A + 1
—— =a
×+1
(3a + 1) divided by (× + 1) = a has no solution

Equilateral triangle. Multiply 2 on the left and right sides of the equation you give you, there is 2 * a ^ 2 + 2 * B ^ 2 + 2 * C ^ 2-2ab-2bc-2ac = 0, that is, (a-b) ^ 2 + (B-C) ^ 2 + (A-C) ^ 2 = 0, obviously a = b = C
Why is there a second problem? Buy one and get one free? The answer to the second problem is a = 0 or 1 / 3. The method is to solve x inversely. X can't take - 1. In this way, we can get a value of a to make the equation have no solution. In addition, X inversely solved contains a as the denominator, so we can get the above results

A problem in factorization should be very simple. I taught myself factorization,
If a × B = m, C × d = q and AC + BD = P, then the polynomial can be factorized as (AX + D) (BX + C)
This passage is quoted from "Twelve methods of factorization (sorted)" in Baidu Library. When I look at this passage, I don't understand that it must be "a × B = m, C × d = q and AC + BD = P". If so, what formula theorem or method is there to constrain the values of M, P and Q? Or an example of the author, which has no other meaning?

M p q is any three numbers without any constraints
If the MPQ in the specific topic does not satisfy the formula you listed (if a × B = m, C × d = q and AC + BD = P), then it can be said that this equation can not be factorized, not to factorize, but to make MPQ conform to a specific constraint condition

If a ^ 2 + A + 1 + 0, (1) find the value of a ^ 3 + 2A ^ 2 + 2A + 1, (2) find the value of a ^ 2001 + A ^ 2002 +. + A ^ 2209
In one question
3/4x^2y-9/8xy^2

A ^ 3 + 2A ^ 2 + 2A + 1 = (a ^ 3 + A ^ 2 + a) + (a ^ 2 + A + 1) = a (a ^ 2 + A + 1) + (a ^ 2 + A + 1) = a * 0 + 0 = 0. It should be a ^ 2001 + A ^ 2002 +. + A ^ 2009a ^ 2001 + A ^ 2002 +. + A ^ 2009 = a ^ 2001 (a ^ 2 + A + 1) + A ^ 2004 + +a^2009=0+a^2004(a^2+a+1)+a^2007+a^2008+a^2009=0+...