if x^2-2011x+1=0 is true,then waht is the value of x-2010x+2011/(x^2+1)?

x² = 2011x - 1  →  x² + 1 = 2011x

So

x^2-2010x+【2011/(x^2+1)] =

[2011x - 1 ] - 2010x + 2011/(2011x) =

x - 1 + 1/x =

[x² - 1x + 1] / x =

[x² + 1 - x ] /x =

[2011x - x] / x =

2010x / x =

2010      (with the restriction that x ≠ 0)

$$\\If\;\; x^2-2011x+1=0 $ is true, then what is the value of $\\\\ x-2010x+\frac{2011}{(x^2+1)}\;\;?\\\\ $first$\\ x^2-2011x+1=0\\ x^2+1=2011x\\\\ $now$\\\\\ x-2010x+\frac{2011}{(x^2+1)}\\\\ =x-2010x+\frac{2011}{(2011x)}\\\\ =x-2010x+\frac{1}{x}\\\\ =-2009x+\frac{1}{x}\\\\$$

Maybe you needed more brackets in your question?  I think that you probably did.  

oh. i am sorry the second equation should be x^2-2010x+2011/(x^2+1)

Maybe you really mean this ??

 (x^2-2010x+2011) /(x^2+1)

Is that right?

x^2-2010x+【2011/(x^2+1)]

$$\\If\;\; x^2-2011x+1=0 $ is true, then what is the value of $\\\\ x-2010x+\frac{2011}{(x^2+1)}\;\;?\\\\ $first$\\ x^2-2011x+1=0\\ x^2+1=2011x\\\\ $now$\\\\\ x^2-2010x+\frac{2011}{(x^2+1)}\\\\ =x^2-2010x+\frac{2011}{(2011x)}\\\\ =x^2-2010x+\frac{1}{x}\\\\ =(x^2+1)-1-2010x+\frac{1}{x}\\\\ =2011x-1-2010x+\frac{1}{x}\\\\ =x-1+\frac{1}{x}\\\\$$

$$\\=\frac{x^2-x+1}{x}\\\\ =\frac{1011x-x}{x}\\\\ =\frac{1010x}{x}\\\\ =1010$$

I think I might have stuffed up somewhere.  Plus I have a reputation for doing things the LONG way.

I'm going to look over it some more.   

No I think i am happy enough with that answer.   

Melody. Lol! That was so funny.

I am glad that you find me entertaining Dragon.  

I thought about the restriction Chris but I didn't put it there because the first statement would not be true if x=0.

But like always - yours is shorter.

I started mine from the previous result where the question was wrong.

That is my excuse and I am sticking with it!  LOL  

I'll put it down to your "left-handedness".............LOL!!!!!

(x^2-2011x+1)/x=0/x

x-2011+1/x=0

x+1/x=2011

x-1+1/x=2011-1=2010

Good job.Is it funny or not to do problem like this?if yes, i will put a problem everyday to the forum 

It IS fun, quinn.......just don't make them too difficult......otherwise, it makes us look bad........

P.S  ....I even had to "cheat" off Melody a bit to get that problem correct!!!!  (Shhh!!!...don't tell her)

Yes I had fun with this problem quinn.

But it is better if you get the question right in the first place.  I liked the Extra brackets that you put in second time round, then we know we have not misinterpreted.

I have included this problem in the next "End of Day Wrap" that i  write.  I invite you to read my wrap.  I write one most nights.

Here is the address of yesterday's one.  They are kept in with the sticky topics.  :)

http://web2.0calc.com/questions/all-questions-answered-to-this-point#r294

(I don't usually include that many questions but I had not written one for a few days)

We really DID LIKE this question Quinn, you are welcome here any time