The process of x2-4x + 3 / x2-5x + 6 + s2-2x + 1 / x2-4x + 3 + 1 / x2-3x-4

The process of x2-4x + 3 / x2-5x + 6 + s2-2x + 1 / x2-4x + 3 + 1 / x2-3x-4

X2-4x + 3 / x2-5x + 6 + s2-2x + 1 / x2-4x + 3 + 1 / x2-3x-4
=(x²-5x+6)/(x²-4x+3)+(x²-4x+3)/(x²-2x+1)-(x²-3x-4)/(1-x²)
=(x-2)(x-3)/(x-1)(x-3)+(x-1)(x-3)/(x-1)²-(x-4)(x+1)/(1+x)(1-x)
=(x-2)/(x-1)+(x-3)/(x-1)+(x-4)/(x-1)
=[x-2+x-3+x-4]/(x-1)
=(3x-9)/(x-1)
X2-4x + 3 of x2-5x + 6 + x2-2x + 1 of x2-4x + 3 + 1 of x2-3x-4
=(x-3)(x-2)/[(x-3)(x-1)]+(x-3)(x-1)/[(x-1)²]+(x-4)(x+1)/[(1+x)(1-x)]
=(x-2)/(x-1)+(x-3)/(x-1)-(x+4)/(x-1)
=x-2+x-3-x-4
=x-9
Then the average of ax1 + B, AX2 + B, axn + B '= (ax1 + B + AX2 + B + + axn + B takes 3x1-2 as a, and because the variance is one third, then 3x2-2 = a + 1, 3x3-3 = a + 2)
Calculation: x2-5x + 6 / x2-4x + 3 + x2-4x + 3 / x2-2x + 1 + x2-3x-4 / 1-x2
x2-5x+6/x2-4x+3 + x2-4x+3/x2-2x+1 + x2-3x-4/1-x2 =(x-2)(x-3)/(x-3)(x-1)+(x-3)(x-1)/(x-1)²-(x-4)(x+1)/(x+1)(x-1)=(x-2)/(x-1)+(x-3)/(x-1)-(x-4)/(x-1)=(x-2+x-3-x+4)/(x-1)=(x-1)/(x-1)=1
General solution of higher order equation of one variable
Is there a general solution to higher order equation of one variable? If so, what is it?
Try to reduce the frequency
2. Only approximate solution can be obtained
Solving equation 2x & # 178; + x = 0 by collocation method
2x²+x=0
2（x²+2×1/4 x+1/16）=1/8
2（x+1/4）²=1/8
（x+1/4）²=1/16
x+1/4=1/4 x+1/4= -1/4
x1=0 x2= -1/2
The solution of higher order equation,
For example, for a binomial equation (x ^ n = a), all roots can be obtained directly. For a general equation of degree five or more, there is no root formula, but for a real coefficient equation, there is no root formula
The equation is solved by the collocation method: (1) x & # 178; - 1 = - x (2) 2x & # 178; - 6x + 2 = 0
（1）x²-1=-x
x²﹣x﹣1=0
x²﹣2×﹙1/2﹚x﹢1/4﹣1/4﹣1=0
﹙x﹣1/2﹚²=﹙√5/2﹚²
x﹣1/2=±√5/2
∴x=﹙1±√5﹚/2
（2）2x²-6x+2=0
x²-x+1=0
﹙x﹣1/2﹚²=﹣3/4
﹙x﹣1/2﹚²≧0
There is no real root
Formula method of quadratic equation with one variable (plus points)
Solve the following equation by formula
X square + 2x-2 = 0 x square + 5x = 7
3x square + 5 * (2x + 1) = 0 root 2x square + 4 times root 3x-2 times root 2 = 0
Given that the root of the equation x square + kx-6 = 0 is 2, find its other root and the value of K
X square + 2x-2 = 0. A = 1, B = 2, C = - 2
x=【-2±√（4+4*2）】/2=-1±√3
X square + 5x = 7. A = 1, B = 5, C = - 7
x=【-5±√53】/2
3x square + 5 * (2x + 1) = 0. A = 3, B = 10, C = 5 are substituted into the formula OK
Root sign 2x square + 4 times root sign 3x-2 times root sign 2 = 0... A = √ 2, B = 4 √ 3, C = - 2 √ 2, substitute into the formula OK
2 into x square + kx-6 = 0. K = 1
X square + X-6 = 0
(X-2) * (x + 3) = 0. The other root is 3
Question 1: X1 = - 1 plus minus root 3
X square + 2x-2 = 0..... A = 1, B = 2, C = - 2
x=【-2±√（4+4*2）】/2=-1±√3
X square + 2x = 0 x square + 5x = 7
X square + 5x-7 = 0
X square = 2x x square + 5x + 5-7 of 2-2 = 0
The square of x = 2 (x + 5 / 2) - 4.5 = 0
... unfold
X square + 2x = 0 x square + 5x = 7
X square + 5x-7 = 0
X square = 2x x square + 5x + 5-7 of 2-2 = 0
The square of x = 2 (x + 5 / 2) - 4.5 = 0
X + 2.5 + radical 4.5) (x + 2.5 - radical 4.5) = 0
X1 = - 2.5 - radical 4.5 x2 = - 2.5 + radical 4.5
Solving the equation by collocation method: 2x2-4x + 1 = 0
The original equation is changed into formula x2 − 2x = − 12, and the formula is x2 − 2x + 1 = 1 − 12, that is, (x − 1) 2 = 12, and the formula is x − 1 = ± 22x = 1 ± 22 = 2 ± 22 ℅ X1 = 2 + 22, X2 = 2 − 22
For a project, it takes two more days for Party A to do it alone than Party B. now Party A has done it alone for three days, and Party B has done it alone for four days. Then 80% of the work has been completed. How many days does it take for Party A and Party B to do it alone?
For a project, it takes two more days for Party A to do it alone than Party B. now Party A has done it alone for three days, and Party B has done it alone for four days. Then 80% of the work has been completed. How many days does it take for Party A and Party B to do it alone?
Let B do X alone, and a do X + 2
3/[X+2]+4/X=4/5
X=8
A: it takes 8 + 2 = 10 days for a to do it alone, and 8 days for B
If it takes X days for a to complete it alone, then it takes (X-2) days for B to complete it alone.
Then a completes 1 / X every day and B completes [1 / (X-2)] every day
3*（1/X）+4*[1/（X-2）]=80%
[3（X-2）+4X]/[X*（X-2）]=4/5
The final solution is: X1 = 40, X2 = 3
Since Party A has done the project for three days and Party B has done it for four days, 80% of the work has been completed. Therefore, the solution of X2 is wrong.
So a does it alone
If it takes X days for a to complete it alone, then it takes (X-2) days for B to complete it alone.
Then a completes 1 / X every day and B completes [1 / (X-2)] every day
3*（1/X）+4*[1/（X-2）]=80%
[3（X-2）+4X]/[X*（X-2）]=4/5
The final solution is: X1 = 40, X2 = 3
Since Party A has done the project for three days and Party B has done it for four days, 80% of the work has been completed. Therefore, the solution of X2 is wrong.
So it takes 40 days for a to do it alone, and 38 days for B to do it alone
When the equation 2x & # 178; - 7x & # 178; = 5 is solved by formula method, the value of ABC is determined first?
A a -7,b 2,c 5
B a 7,b 2,c -5
C a 7,b -2,c 5
D a 7,b -2,c -5
2x-7x²=5
7x²-2x+5=0
So it is
C a 7,b -2,c 5
Have you entered the wrong title
transposition:
2x-7x & # 178; - 5 = 0 times - 1:7x & # 178; - 2x + 5 = 0
So a 7, B - 2, C 5
Choose C
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