The roots of the two real numbers x 2 x 1 0 are x1 and x2 Find the value of 1 x1 10 x2 8 2 x1 80

Since x1,x2 is the root of the equation x2-x-1=0, x1+x2=1,x1x2=-1

From x 2-x-1=0 we get x 2=x+1, so x 4=x 2+2x+1=x+1+2x+1=3x+2x 8=9x 2+12x+4=21x+13x 10=(21x+13)(x+1)=55x+34, so x1 10=55x1+34, x2 8=21x2+13, so x1 10+x2 8=55x1+34+21x2+13=34x1+21(x1+x2)+47=34x1+68

2) From x1x2=-1 we get 1 x2=-x1, so x1 806+1 x2 10=x1 806+(-x1) 10=x1 806+x1 10

The calculation is an astronomical number, and the question is not copied incorrectly.

The root of the original equation x1=(1-5) 2,x2=(1+ 5) 2 The nth power of the root of the original equation can be expressed by the first square: x n=a(n-1)x+a(n-2), a(n) is the general term of the Fipochna sequence, a(n)=[((1+ 5) 2) (n+1)-(1- 5) 2) (n+1)] 5, n,is an integer.

The first 11 terms of the sequence (n=0 to 10) are as follows: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, and the characteristic of this sequence is that from the third term onwards, each term is the sum of the first two terms, and the Fipochna sequence is a series of rational numbers expressed in irrational numbers.

1) (x1) 10=a(9)x1+a(8)=55x1+34, (x2) 8=a(7)x2+a(6)=21x2+13, so (x1) 10+(x2) 8=55x1+21x2+47=85-17 5. (2)1/(x2)^10=(x1)^10=(55-55√5)/2；In addition, due to |x1|,a(806) 3 10 168,so(x1) 806 0,so(x1) 806+1 (x2) 10 (x1) 10=(55-55 5) 2(bi).

According to the relationship between the root and the coefficient, we can know that x1+x2=4 x1x2=1 Then simplify the required formula and directly bring the two formulas into the number, no need to find x1 and x2 respectively (because the process is not very easy to play, you can do it, don't ask me again).

By the formula a n-b n=(a-b)(a (n-1) + a (n-2)*b....b^(n-1))

Then you can know that the above formula can be simplified and hungry.

x^100-1

Bring x=2 in to get the calculated crack formula.

The answer is 2 100-1

Answer: Solution: x1, x2 are the two real roots of the equation x2+x-1=0, x1+x2=-1;

and x13=x1x12

x1（1-x1）

x1-x12

2x1-1-2x22

2（1-x2）

2+2x2，x13-2x22+2008

2x1-1-2+2x2+2008

2（x1+x2）+2005

So the answer is: 2003

This problem mainly examines the relationship between roots and coefficients, algebraic evaluation Combining the relationship between roots and coefficients with algebraic deformation is a frequently used method of problem solving

x1, x2 is the root of the equation x x 1=0, then: (x1) 2x1) 1=0 is:

x1)²=1－x1。In the same way, there is: (x2) = 1 (x2), and there is: x1 x2 = 1

x1)³－2(x2)²＋2008

x1)[1－x1]－2(1－x2)＋2008=(x1)－(x1)²－2＋2(x2)＋2008=x1－[1－x1]＋2x2＋2006

2x1＋2x2＋2005

2(x1＋x2)＋2005

From the question: x = 1-x, for x1 and x2 are true.

Then: x1 = x1· 1 x1 = x1 x1 = x1 (1 x1) = 2·x1 1

2·x2 = 2(1 x2) = 2 2·x2 = 2·x1 1 ( 2 2·x2) 2008=2(x1+x2)+2005

This is an examination of the nature of the root of equations and Veda's theorem.

Observing (x1) 2-3x1-x2 finds that it can be split into [(x1) 2-2x1]-(x1+x2), which is known by x1 as the root of the equation [(x1) 2-2x1]=1 by Veda's theorem (x1+x2)=2

So (x1) 2-3x1-x2

(x1)^2-2x1]-(x1+x2)=1-2=-1

This is actually not a difficult question.

Is it possible to dissolve x1 and x2 from the first equation and substitute equation 2 for equation 2?

- The original equation has x1 = 1 + root number 2 x2 = 1 - root number 2

Using Veda's theorem, the original formula = x1*x2-(x1+x2)+1=-1-2+1=-2

Did Vida theorem? x1x2=c a,x1+x2=-(b a), you can group the desired formula. Give it a try!

The solution x1 and x2 are the root of the equation.

There is Veda's theorem

x1+x2=4

x1x2=1

x1+x2)²÷1/x1+1/x2)=16÷[(x1+x2)/(x1x2]

Original Trillion Liquid Call x1x2(x1+x2)-x1x2 and x1+x2=-b a=-99

x1x2=a/c=-1

Buried in the clan Kaiyuan -1x(-99)-(1).