Knowing x x 1 0, find x 1 2 1 x x 1 x 1 x 1 x 2x 1

First, find the value of x, which is equal to (-1 + root number 5) 2 (-1 - root number 5) 2

Then calculate. Always, right?

So, I don't have much time to take my time? ok？

Solution: x(1-2 1-x) (x+1)-x(x -1) x -2x+1x(1-2 1-x) (x+1)-x(x -1) (x-1)(x-1).

x-x²-2x)÷(1-x)(x+1)-x(x+1)(x-1)/(x-1)(x-1)

x-x²)÷1-x)(x+1)-x(x+1)/(x-1)x(1+x)÷(x-1)(x+1)-x(x+1)/(x-1)x÷(x-1)-x(x+1)/(x-1)

x-1)x-x²-x)/(x-1)

x²)/(x-1)

x²/(1-x)

Where the known x +x-1=0 can be transformed into x =1-x, so the equation answer = 1

Simplified: x (x-1)-(x3-x) (x-1)2 passes are divided into: x2(i-x) (x-i)2

Resimplify: x2 (1-x).

x2=1-x

So Answer 1

I wouldn't square x, so write x2

x -3x+1=0, because x is not 0, so the two sides are divided by x to obtain:

x－3＋1／x＝0

i.e. x+1 x 3

squared: x 2+1 x 2+2=9

x²+1/x²=9-2=7

I don't understand the formula for solving. Can you write carefully?

The equation is finally reduced to -1 (x-1).

From the known x=(-1 root number 5) 2, so the value of the final equation is 2 (2 root number 5).

Is there a mistake in the question?

The value of x can be found according to the root finding formula: [1+sqr(5)] 2 or [1-sqr(5)] 2. In this way, the expression that follows can be simplified and the data can be substituted.

Can you make the brackets a little bigger? Otherwise, it's hard to see which is the denominator and which is the numerator in (x-1, x-x-2, x+1) (2x-x x +2x+1)?

From x -3x-1=0 we get x-1 x=3

The squares of the two sides give x +(1 x) -2=9, x +(1 x) =11.

x²-(1/x)²]=[x²+(1/x)²]4=117，∴x²-(1/x)²=±3√13

Divide the known equation by x at the same time, and get x-1 x=3

x²+1/x²=(x-1/x)²-2=7

You're on the wrong topic.

Find x from the previous formula, and then substitute it for the next formula, which is a bit troublesome, but it is easy to use.