You can only have two terms in each "frac" (I still use dfrac myself but it appears it's much the same thing now)
[tex]\frac{2x}{x^2(2x+1)}[/tex] =
[tex]\lim_{x \to \infty} \left(1+ \dfrac{1}{x}\right)^x = e[/tex] is
If you want to test your latex you can create a thread in http://www.mathhelpforum.com/math-help/f47/ if you title it "test"
I'll put down what I have done but can't get it right?
2x(x+1) + 3(x+1)(2x - 3) - 2x(2x - 3) = in stages;
2x(x+1) = 2x^2 + 2x
3(x+1)(2x-3) = 2x^2 - 3x + 2x - 3
3(2x^2 - 5x - 3) = 6x^2 -15x + 9
- 2x(2x - 3) = - 4x^2 + 6
combine terms
2x^2 + 2x + 6x^2 -15x + 9 + - 4x^2 + 6 = 8x^2 - 13x + 9
then the denominator;
(2x - 3)(x+1) = 2x^2 + 2x - 3x - 3 = 2x^2 -x - 3
The result I get;
8x^2 - 13x + 9 / 2x^2 -x - 3
4x^2 + x + 1
But I am not right?
Why are you still expanding the denominator? You should leave the denominator factored, as we advised previously.
is where you start to get confused.Originally Posted by David
I suggested writing this as in a previous reply also. Then, when you foil it, you get
This is the only mistake I can see, though - the rest was very good.
[1]
Note your signs: . Consequently the next line has a carried error and should read
[2]
You need to distribute the x on the second term: instead of just 6
[3]
Look at your coefficients of x^2.
The numerator, once expanded is below. I've put in brackets to show each term for clarity's sake. They are not mandatory.
Collecting like terms:
Thus your numerator is
Please advise so I understand why is 3(x+1)(2x-3) worked out as;
2x^2 - 3x + 2x - 3 = 2x^2 -x - 3
3(2x^2 -x -3) = 6x^2 -3x +9 wrong
I don't understand how I would know that I should put the 3(x+1) inside the bracket as (3x +1) because working with it the results are different.
also in your example; (3x + 3)(2x - 3) = 6x^2 + 6x - 9x - 9
then I get (3x + 3)(2x - 3) = 6x^2 - 9 + 6x - 9
I assume then that you tidy the result up as 6x^2 + 6x -9x -9
Thanks
David
Only the sign on 9: it should be
You put in (3x+3) in the bracket. Personally I prefer to leave multiplying by 3 until last but both methods are good.I don't understand how I would know that I should put the 3(x+1) inside the bracket as (3x +1) because working with it the results are different.
You can further simplify 6x-9x to -3xalso in your example; (3x + 3)(2x - 3) = 6x^2 + 6x - 9x - 9
then I get (3x + 3)(2x - 3) = 6x^2 - 9 + 6x - 9
I assume then that you tidy the result up as 6x^2 + 6x -9x -9
Thanks
David