5x/1*2+5x/2*3+5x/3x4+5x/4*5 +.+5x/2003*2004+5x/2004*2005=1

5x/1*2+5x/2*3+5x/3x4+5x/4*5 +.+5x/2003*2004+5x/2004*2005=1

5x/1*2=5x（1-1/2） 5x/2*3=5x（1/2-1/3）
5x/1*2+5x/2*3+5x/3x4+5x/4*5 +.+5x/2003*2004+5x/2004*2005=5x（1-1/2+1/2-1/3.-1/2005）=5x（1-1/2005）=1
x=401/2004
Solution equation: 5x / (1 * 2) 5x / (2 * 3) 5x / (3 * 4) 5x / (4 * 5)... 5x / (2003 * 2004) 5x / (2004 * 2005) = 1
The solution equation: 5x / (1 * 2) + 5x / (2 * 3) + 5x / (3 * 4) + 5x / (4 * 5) +... + 5x / (2003 * 2004) + 5x / (2004 * 2005) = 1.5x (1 / 1-1 / 2 + 1 / 2-1 / 3 + 1 / 3-1 / 4 + 1 / 4-1 / 5 + +1 / 2003-1 / 2004 + 1 / 2004-1 / 2005) = 1,5x (1-1 / 2005) = 1,5x (2004 / 2005) = 1
In order to expand sales, reduce inventory and increase profits, the department store decided to take appropriate price reduction measures. After market research, it was found that if the price of each children's clothing was reduced by 1 yuan, 2 more children's clothing could be sold every day. If we want to make an average profit of 1 yuan every day in selling this kind of children's clothing 200 yuan, then how much should each children's clothing be reduced?
Suppose that every piece of children's clothing should be reduced by X Yuan. From the meaning of the question: (40-x) (20 + 2x) = 1200, the solution is: x = 10 or x = 20
Using the collocation method to solve the equation: 3x & # 178; - X-1 = 0
3x & # 178; - X-1 = 0, divide both sides by 3, get X & # 178; - (1 / 3) x - (1 / 3) = 0 shift, get X & # 178; - (1 / 3) x = (1 / 3) add (1 / 6) &# 178;, get X & # 178; - (1 / 3) x + (1 / 6) &# 178; = (1 / 3) + (1 / 6) &# 178; [x - (1 / 6)] &# 178; = (13 / 36) square of both sides, get X - (1 / 6) = (± √ 13) / 6x
We know the equation x 2 - (2m + 1) x + m 2-4 = 0 about X. if the sum of squares of two unequal real roots of the equation is equal to 15, we can find the value of M
According to the meaning of the question, we get △ = (2m-1) 2-4 (M2-4) > 0, and the solution is m ＞ - 174. Let two of the equations be a and B, then a + B = 2m + 1, ab = M2-4, ∵ A2 + B2 = 15, ∵ (a + b) 2-2ab = 15, ∵ (2m + 1) 2-2 (M2-4) = 15, and we get M2 + 2m-3 = 0, and M1 = - 3, M2 = 1, ∵ m is - 3 or 1
Solving the equation by 4 & # 178; - 3x-1 = 0 collocation method
4(x^2-3/4x+(3/8)^2)-4*(3/8)^2-1=0
(x-3/8)^2=[4*(3/8)^2+1]/4
x-3/8=±5/2
X = 23 / 8 or - 17 / 8
What is the solution to higher order equation?
Is the solution of cubic equation of one variable OK?
The solution of the root formula of cubic equation of one variable
-------From high school mathematics website
The formula for finding roots of cubic equation with one variable can not be made by ordinary deductive thinking. The matching method similar to the formula for finding roots of quadratic equation with one variable can only formalize the standard cubic equation with one variable of type ax ^ 3 + BX ^ 2 + CX + D + 0 into a special type of x ^ 3 + PX + q = 0
The solution of the formula for solving the cubic equation of one variable can only be obtained by inductive thinking, that is, the form of the formula for finding the root of cubic equation of one variable can be concluded according to the form of the formula for finding the root of linear equation of one variable, quadratic equation of one variable and special higher order equation. The form of the formula for finding the root of cubic equation of one variable, such as x ^ 3 + PX + q = 0, should be x = a ^ (1 / 3) + B ^ (1 / 3), It is the sum of two open cubes. The form of the formula for finding the root of cubic equation with one variable is summed up. The next step is to find out the content in the open cube, that is, to use P and Q to express a and B
(1) Let x = a ^ (1 / 3) + B ^ (1 / 3) be cubic at the same time
（2）x^3=(A+B)+3(AB)^(1/3)(A^(1/3)+B^(1/3))
(3) Because x = a ^ (1 / 3) + B ^ (1 / 3), so (2) can be changed into
X ^ 3 = (a + b) + 3 (AB) ^ (1 / 3) X
(4) Compared with the univariate cubic equation and the special type of x ^ 3 + PX + q = 0, we can see that x ^ 3-3 (AB) ^ (1 / 3) x - (a + b) = 0
(5) - 3 (AB) ^ (1 / 3) = P, - (a + b) = Q
（6）A+B＝－q,AB＝-（p/3）^3
(7) In this way, the root formula of cubic equation with one variable is transformed into the root formula of quadratic equation with one variable, because a and B can be regarded as the two roots of quadratic equation with one variable, and (6) is the Veda theorem about the two roots of quadratic equation with one variable in the form of ay ^ 2 + by + C = 0
（8）y1＋y2＝－（b/a）,y1*y2=c/a
(9) Comparing (6) and (8), we can make a = Y1, B = Y2, q = B / A, - (P / 3) ^ 3 = C / A
(10) Because the root formula of quadratic equation of type ay ^ 2 + by + C = 0 is
y1＝－（b＋（b^2－4ac）^(1/2)）/(2a)
y2＝－（b－（b^2－4ac）^(1/2)）/(2a)
Can be transformed into
(11)y1＝－(b/2a)-((b/2a)^2-(c/a))^(1/2)
y2＝－(b/2a)+((b/2a)^2-(c/a))^(1/2)
Substituting a = Y1, B = Y2, q = B / A, (P / 3) ^ 3 = C / A in (9) into (11), we can get
(12)A＝－(q/2)-((q/2)^2＋（p/3）^3）^(1/2)
B＝－(q/2)+((q/2)^2＋（p/3）^3）^(1/2)
(13) Substituting a and B into x = a ^ (1 / 3) + B ^ (1 / 3), we get
(14)x=（－(q/2)-((q/2)^2＋（p/3）^3）^(1/2)）^(1/3)+（－(q/2)+((q/2)^2＋（p/3）^3）^(1/2)）^(1/3)
Equation (14) is only a real root solution of one variable cubic equation. According to Weida's theorem, one variable cubic equation should have three roots, but according to Weida's theorem, as long as one of the roots is solved, the other two roots can be easily solved
X ^ y is the Y power of X
It's very complicated
The solution of cubic equation with one variable discovered by tatalia
The general form of cubic equation of one variable is
x3+sx2+tx+u=0
If we make an abscissa translation y = x + S / 3, then we can eliminate the quadratic term of the equation
Go. So we just have to think about form
x3=px+q
The cubic equation of the equation
Suppose that the solution X of the equation can be written as x = A-B, where a and B are undetermined parameters
If we plug in the equation, we have
a3-3a2b+3ab2-b3=p(a-b)+q
Sort it out
a3-b3 =(a-b)(p+3ab)+q
According to the theory of quadratic equation, we can select a and B appropriately, so that when x = A-B,
3AB + P = 0
a3-b3=q
Multiply each side by 27a3 to get
27a6-27a3b3=27qa3
It can be seen from P = - 3AB
27a6 + p3 = 27qa3
This is a quadratic equation about A3, so we can solve A. then we can solve B and root X
Solution of quartic equation of one variable discovered by Ferrari
As in the cubic equation, the quartic equation can be eliminated by a coordinate translation
Therefore, as long as we consider the following form of quaternion equation with one variable:
x4=px2+qx+r
The key is to use parameters to match the two sides of the equation into a complete square form. Consider a parameter
a. We have
(x2+a)2 = (p+2a)x2+qx+r+a2
On the right side of the equation is a complete square if and only if its discriminant is 0
q2 = 4(p+2a)(r+a2)
This is a cubic equation about A. using the solution of the cubic equation with one variable above, we can
So that both sides of the original equation are completely squared, and the root of the square is a function of X
Then the root X of the original equation can be solved
Finally, there is no general algebraic solution for the equation of higher degree of one variable of degree 5 or more (that is, through the coefficients through the finite four arithmetic operation and the operation of power and root), which is called Abel's theorem
http://www.xycq.net/forum/archiver/?tid-85077.html
http://www.hbedu.com.cn/2006-2-7/20062781401.htm
http://www.wlck.com/bbs/printpage.asp?BoardID=32&ID=6599
These three websites are the solution of the equation of one variable quartic!
Y = - 3x & # 178; + 3x + 1 use the collocation method to solve the equation
Y=-3(x²-X+1/4)+7/4
Y=-3(X-1/2)²+7/4
X=1/2
Y=7/4
The solution of higher order equation of one variable
a1x^n+a2x^(n-1)+…… +（an）x+（an+1）=0
The method of finding approximate solution is as follows
1. Newton iterative formula;
2. Dichotomy;
Note: it is usually solved by procedure;
Solving equations by 2x & # 178; - X-1 = 0 collocation method
2x²-x-1=0
x²-0.5x=0.5
x-0.5x+0.0625=0.5625
(x-0.25)²=0.5625
x-0.25=±0.75
x=0.25±0.75
x1=-0.5
x2=1