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Substituting the five steps of elimination method, adding and subtracting the five steps of elimination method
Addition and subtraction elimination method should be 4 steps
1. Partition coefficient. The coefficient of one element of two equations is the same
2. Elimination. Two equations are added or subtracted to eliminate one element
3. Simplify. Simplify the equation obtained in 2 and get a solution
4. Bring in. Bring the solution of 3 into one of the two original equations and get the other solution
The concept, procedure and method of elimination method for solving binary linear equations
1、 Conceptual steps and methods:
1. From an equation in a system of linear equations of two variables, an unknown number is expressed by an equation containing another unknown number, and then it is substituted into another equation to realize elimination, and then the solution of the system of linear equations of two variables is obtained. This method is called substitution elimination method for short
2. The steps of solving binary linear equations by substitution elimination method
(1) A simple equation is selected from the equations, and one of the unknowns is expressed by an equation containing another
(2) The equation obtained in (1) is substituted into another equation to eliminate an unknown number
(3) The value of an unknown number can be obtained by solving the linear equation of one variable
(4) Substituting the value of one unknown into the equation obtained in (1), the value of another unknown is obtained, and the solution of the equation system is determined
Note: (1) when using the substitution method, one equation must be transformed into another equation, otherwise the form of "0 = 0" will be obtained, and the value of the unknown can not be obtained
(2) when the coefficient of an unknown number of an equation is 1 or - 1, the substitution method is more convenient
3. When the coefficients of the same unknowns in two bivariate linear equations are opposite or equal, the two sides of the two equations can be added or subtracted respectively to eliminate the unknowns and obtain a bivariate linear equation. This method is called addition subtraction elimination method
The basic idea of solving binary linear equations by addition subtraction elimination method is still elimination
4. The general steps of solving binary linear equations by addition and subtraction method: Step 1: if the coefficients of an unknown number are opposite to each other, the two sides of the two equations can be added respectively to eliminate the unknown number; if the coefficients of the unknown number are equal, the two sides of the two equations can be directly subtracted, The second step: if the absolute values of the coefficients of an unknown number do not exist in the equations are equal, then a group of coefficients should be selected (a group of coefficients with smaller least common multiple should be selected), their least common multiple should be calculated (if a coefficient is an integral multiple of another coefficient, the coefficient is the least common multiple), and then the original equations should be deformed, Make the absolute values of the coefficients of the new equations equal (all equal to the least common multiple of the original coefficients), and then add, subtract and eliminate the elements. Step 3: for the more complex binary linear equations, we should first simplify them (remove the denominator, remove the brackets, merge the similar terms, etc.), and usually make each equation in the form of the term containing the unknown number on the left side of the equation and the constant term on the right side of the equation, and then consider the above addition, subtraction and elimination
Note: (1) when the absolute values of the coefficients of the same unknowns in two equations are equal or integral multiples, the addition and subtraction method is simpler
(2) if the given equations are complex and difficult to observe, we should deform them first (remove denominator, bracket, shift term, merge, etc.), and then judge which method is better
5. Set up equations to solve simple practical problems. The key to solve practical problems is to understand the meaning of the problem and find out the equal relationship between quantities. The equal relationship here should be two or three. Correctly list one (or several) equations and then form the equations
6. The general steps of solving practical problems with binary linear equations:
(1) set two unknowns in the question;
Using addition subtraction elimination method to solve bivariate linear equation
1、4x-3y=-2
x/4+y/3=3
2、2（m+1）=n-3
4（n-4）=3（m+3
1. 4x-3y=-2 （1）
If x / 4 + Y / 3 = 3 (2) (2) * 16, 4x + 16 / 3 * y = 48 (3)
(3) (1) 25 / 3 * y = 50, y = 6, x = 4
2. It can be reduced to 2m-n = - 5 (1)
4n-3m=25(2)
(1) * 4 + (2) get 5m = 5, M = 1, n = - 7
Solving the quadratic equation of two variables
{2x+3y=0
{4x-9y+1=0
2x+3y=0 -------------(1)
4x-9y+1=0 -----------(2)
(1) * 3
6x+9y=0 --------------(3)
(2) + 3
10x+1=0
X = - 1 / 10
Substituting (1) gives y = 1 / 15
|X-1 | + | X-2 | + | x-3 | + | x-4 | + | X-5 |, when x = 3, the minimum value is 6
1. Five workbenches ABCDE are arranged in sequence on the work line. Where can a toolbox be placed to make the sum of the distances taken by the operators on the workbenches shortest?
2. If the worktable is changed from five to six, then how should the toolbox be placed so that the sum of the six operators can go to work is the shortest?
3. When there are n (n is an integer) worktables on the pipeline, how to place the toolbox most appropriately?
Is the assembly line straight?
If it's a straight line, put the tool in the middle. The first question is C
If the permutation is ABCDEF, put C or D
The third question is n / 2 (round up) (whether from the beginning or from the end)
To explain, your title perfectly explains the first question
If the worktable is not necessarily a straight line, then put it into a circle, tool center!
Factorization: 2 2007-2 2006-2 2005 - -1
Can we do this: turn the whole formula upside down to [(- 1) + (- 2) + +（-22006）]+22007
The answer in brackets can be regarded as an arithmetic sequence. The first n terms and formula can be used to get the answer in brackets. Finally, 22007 is added
Original formula = 22006 / 2 [(- 1) + (- 22006)] + 22007
=-242143021+22007
=242121014
2007x2005 / 2006 + 2005x2007 / 2006 simple operation!
Hello: 2007x2005 / 2006 + 2005x2007 / 2006 = (2006 + 1) x2005 / 2006 + 2005x (1 + 1 / 2006) = 2006x2005 / 2006 + 1x2005 / 2006 + 2005x1 + 2005x1 / 2006 = 2005 + 2005 / 2006 + 2005 / 2006 = 4010 + 4010 / 2006 = 4010 + 1 + 1002 / 1003 = 4011 and 1002 / 1003
148148148 * 2006 simple calculation
You can't use a calculator
The calculation process is given
Don't play those copy and paste tricks!
=148*1001001*(2000+6)
=(148*2000+148*6)*1001001
=(296000+888)*(1000000+1000+1)
=296888*1000000+296888*1000+296888*1
=297185184888
148148148*2006
=(148+148000+148000000*2006
=296888+296888000+296888000000
=297185184888
Remember to give points
How to calculate 2006 times 2006 / 2007?
2006x2006/2007
＝（2007-1）x2006/2007
=2007x2006/2007-2006/2007
=2006-2006/2007
=2005 and 1 / 2007
(x + 1) (x + 2) (x + 3) (x + 4) - 15 (factorization)
Original formula = (X & # 178; + 5x + 4) (X & # 178; + 5x + 6) - 15
Let X & # 178; + 5x = t
The original formula = (T + 4) (T + 6) - 15 = T & # 178; + 10t + 24-15 = T & # 178; + 10t + 9 = (T + 1) (T + 9)
=（x²+5x+1）（x²+5x+9）
Factorization x ^ 15-1
I hope someone will give an answer as soon as possible
a^n-b^n=(a-b)(a^n-1+a^n-2 b+… +ab^n-2+b^n-1)
x^15-1=x^15-1^15=（x-1）（x^14-x^13+x^12-…… -x^3+x^2-x+1)
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