Solving equation by factorization: 25 (X-7) ^ = 16 (x + 4)^

Solving equation by factorization: 25 (X-7) ^ = 16 (x + 4)^

25（x-7）²=16(x+4)²
25（x-7）²-16(x+4)²=0
[5(x-7)+4(x+4)][5(x-7)-4(x+4)]=0
(9x-19)(x-51)=0
x1=19/9,x2=51

If f (x) = ln x + A / 2-x is an odd function, then a=
Why? {[(x+a)/(2-x)][(-x+a)/(2+x)]}=1

Odd function
f(x)+f(-x)=0
ln[(x+a)/(2-x)]+ln[(-x+a)/(2+x)]=0
{[(x+a)/(2-x)][(-x+a)/(2+x)]}=1
(a+x)(a-x)=(2+x)(2-x)
a²-x²=4-x²
a=±2
a=-2
True = - 1, not true
So a = 2

The two roots of equation AX's square + BX + 4 = 0 are 1 and - 3 respectively to find the value of a and B
Square of equation x - (2m + 1) square of X + m + M = 0
If the quadratic equation (1-k) x has two unequal real roots, then the value range of K is ()
A.K>2 B.K

1. The two roots of the equation AX square + BX + 4 = 0 are 1 and - 3 respectively. To find the value of a and B, we can get a = - 4 / 3 B = - 8 / 32 from the solution of Weida's theorem 1 + (- 3) = - B / a 1 * (- 3) = 4 / A. the square of X - (2m + 1) x + m + m + m + m + M = 0 (x-m) (x-m-1) = 0, and get x = m or x = m + 13. (1-k) x's Square - 2x-1 = 0

When a store sells a commodity, it has the following plan: ① increase the price by 10%, and then reduce the price by 10%; ② reduce the price by 10%, and then increase the price by 10%; ③ increase the price by 20%, and then reduce the price by 20%
A. (2) the price adjustment results of the two schemes are the same before and after the B.. Neither of the three plans has recovered the original price. The C. plan has been restored to the original price. The price of the D. scheme is higher than that of the plan.

the original price of this commodity is a yuan, and the price after the price adjustment is: (1+10%) (1-10%) a=0.99a yuan; (1-10%) (1+10%) a=0.99a yuan; (1+20%) (1-20%) a=0.96a element summed up as follows: (1) the price adjustment results of the two schemes are the same before and after, so the right is right; the three option is...

Factorization of a-squared-4ab + 4b-squared-9

Ask an algebra proof problem in the second semester of junior one
It is known that: the square of a + the square of B + the square of C - AB BC CA = 0, and it is proved that: a = b = C

It is proved that a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ca = 0
If you multiply both sides of the equation by two, you get
2a^2+2b^2+2c^2-2ab-2bc-2ca=0
Then, a ^ 2-2ab + B ^ 2 + B ^ 2-2bc + C ^ 2 + A ^ 2-2ca + C ^ 2 = 0
That is: (a-b) ^ 2 + (B-C) ^ 2 + (A-C) ^ 2 = 0
Because: complete square numbers are all non negative numbers, to make the equation hold, only A-B = 0, B-C = 0, a-c = 0
So a = b = C

Two methods of producing surplus value and dialectical relationship?

These two methods are absolute surplus value production and relative surplus value production

A three digit number, if the one digit number and the hundred digit number are exchanged, the three digit number obtained is 693 larger than the original number
For a three digit number, if the single digit number is exchanged with a hundred digit number, the three digit number obtained is 693 larger than the original number; if the ten digit number is exchanged with a hundred digit number, it is 450 larger than the original number. Moreover, the single digit number of the original three digit number is 1 larger than the sum of the hundred digit number and the ten digit number

zyx - xyz = 693
=>z>x
x-z+10=3 (1)
yxz -xyz =450
=>y>x
10+x-y=5 (2)
z=x+y+1 (3)
From 1 x = z-7
From 2 y = 5 + X
Substituting 3
z= z-7 + 5 +z-7 +1
z=8
x=1
y=6
The original number is 168

The application of a mathematical equation of grade one,
The charging standard of a parking lot is as follows: the parking fee for medium-sized cars is 6 yuan / car, and that for small cars is 4 yuan / car. There are 50 small and medium-sized cars in the parking lot, and these cars pay a total parking fee of 230 yuan. How many small and medium-sized cars are there?
We're talking about doing this for Mao

Idea: there are 50 cars in total, assuming that there are x medium-sized cars, then the small cars are (50-x). When you know the number of cars, and then multiply by their respective parking fees, you can work out the equation
If there are x medium-sized cars, then (50-x) for small cars
6x+4（50-x）=230
The solution is x = 15
There are 15 medium-sized cars and 35 small cars

P is any point on the diagonal BD of parallelogram ABCD. The straight line passing through point P intersects ad at point m, BC at point n, the extension line of BA at point E, and the extension line of DC at point F

∵AB//CD
∴PE/PF=PB/PB
∵AD//BC
∴PB/PD=PN/PM
∴PE/PF=PN/PM
That is PE × PM = PF × PN