Junior high school algebra to prove the problem, the use of the ratio of proportion in the theorem Ask the teachers and students of junior high school for a proof of junior high school algebra, It is known that a = B + C + 1; d = e + F + 1; g = H + I + 1; Proof: (D-A) / (D-G) = (E-B) / (E-H) = (F-C) / (f-I) Should be the use of proportion in the split ratio theorem After so many years of reading, I finally forgot all my junior high school mathematics, The recommended answers are: Let Ka = IB + JC + 1 a=b+c+1->k1A=i1B+j1C+1 d=e+f+1->k2A=i2B+j2C+1 G = H + I + 1 - > K3A = i3b + j3c + 1 Substituting (D-A) / (D-G) = (k2-k1) a / (k2-k3) a = (k2-k1) / (k2-k3) Similarly, (E-B) / (E-H) = (F-C) / (f-I) = (k2-k1) / (k2-k3) It has been proved But there are some questions: (E-B) / (E-H) should not be equal to (i2-i1) / (i2-i3), how can it be equal to (k2-k1) / (k2-k3)?

Junior high school algebra to prove the problem, the use of the ratio of proportion in the theorem Ask the teachers and students of junior high school for a proof of junior high school algebra, It is known that a = B + C + 1; d = e + F + 1; g = H + I + 1; Proof: (D-A) / (D-G) = (E-B) / (E-H) = (F-C) / (f-I) Should be the use of proportion in the split ratio theorem After so many years of reading, I finally forgot all my junior high school mathematics, The recommended answers are: Let Ka = IB + JC + 1 a=b+c+1->k1A=i1B+j1C+1 d=e+f+1->k2A=i2B+j2C+1 G = H + I + 1 - > K3A = i3b + j3c + 1 Substituting (D-A) / (D-G) = (k2-k1) a / (k2-k3) a = (k2-k1) / (k2-k3) Similarly, (E-B) / (E-H) = (F-C) / (f-I) = (k2-k1) / (k2-k3) It has been proved But there are some questions: (E-B) / (E-H) should not be equal to (i2-i1) / (i2-i3), how can it be equal to (k2-k1) / (k2-k3)?

It is proved that: (D-A) / (D-G) = [(E + F + 1) - (B + C + 1)] / [(E + F + 1) - (H + I + 1)] = [(E-B) + (F-C)] / [(E-H) + (f-I)] from the theorem of combining ratio (A / b = C / D = = > A / b = (a + C) / (B + D)) we get (D-A) / (D-G) = [(D-A) + (E-B) + (F-C)] / [(D-G) + (E-H) + (f-I)]. Let the value of the above formula be m, that is (D-A) / (...)

Mathematical problems on algebraic application of fractional equation
1. If (4x) / (x * x-4) = A / (x + 2) - B / (X-2), try to find the value of a * a + b * B
2. It is known that a, B and C are all positive numbers, and a / (B + C) = B / (a + b) = C / (a + b) = K
Then the coordinates of the following four points on the image with positive scale function y = KX are
A、（1,1/2） B、（1,2） C、（1,-1/2） D、（1,-1）
Note: please write the division sign as a fraction. I can't mark
The answer must be clear about the process, otherwise I can't understand it, even if I know the answer, it's useless

1:(4x)/(x^2-4)=(4x)/[(x+2)(x-2)]=(4x)[(1/(x+2)-1/(x-2)](-1/4)=
(- x) / (x + 2) - (- x) / (X-2), so a = - x, B = - x, a * a + b * b = 2x * X
2: A = K (B + C), B = K (a + C), C = K (a + b), sum of three formulas
A + B + C = K (B + C + A + C + A + b) = 2K (a + B + C), because a, B and C are all positive numbers,
A + B + C > 0, so 2K = 1, k = 1 / 2, y = KX = x / 2,
So we know that point a is on y = KX

On the method of solving an algebra problem in grade one of junior high school
This problem is as follows: find the natural number (A1A2... An) [this number is the whole], so that 12 times (2a1a2... An1) = 21 times (1a1a2... An2). The answer is to multiply the numbers on both sides of the equal sign, 12 times (2a1a2... An1) = 2 (4 + A1) (2A1 + 2)... (2An + 1) 2. How can it be like this? I use the set value to bring in a set of formulas, but what is the solution?

For example, (123) means 123. Let (A1A2... An) = x (2a1a2... An1) = 2 * 10 ^ (n + 1) + 10 * x + 1 (1a1a1a2... An2) = 1 * 10 (n + 1) + 10x + 2