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8 answers

4!

I knew I could do it.

Damn ma hope yer proud

2006-07-25 21:13:53 · answer #1 · answered by rainphys 2 · 1 2

(2x + 2)/(x+1) + 1/x Set this equal to 0
(2x+2)/(x+1) + 1/x = 0 Now subtract 1/x from both sides
(2x+2)/(x+1) = -1/x Now multiply both side by x
(2x^2+2x)/(x+1) = -1 Multiply both side by x-1
2x^2+2x = -(x+1) Now add x+1 to both sides
2x^2+2x+x+1 = 0 Collect your like terms
2x^2 + 3x +1

2006-07-25 20:49:29 · answer #2 · answered by Anonymous · 0 0

do you mean: [(2x+2/x+1) + 1] / x?
[2(x+1)/(x+1) + 1] / x = (2 + 1) / x = 3/x.

2006-07-25 18:37:24 · answer #3 · answered by the redcuber 6 · 0 0

(2x+2)/(x+1) + 1/x
= [(2x+2)(x) + (x+1)]/(x+1)(x)
=(2x^2+2x + x+1)/(x^2 + x)
=(2x^2+3x+1)/(x^2+x)
=[(2x+1)(x+1)]/[(x)(x+1)]
=2 + 1/x

2006-07-25 18:55:45 · answer #4 · answered by sumone^^ 3 · 0 0

((2x + 2)/(x + 1)) + (1/x)
(2(x + 1)/(x + 1)) + (1/x)
2 + (1/x)
((2x)/x) + (1/x)
(2x + 1)/x

2006-07-26 03:02:54 · answer #5 · answered by Sherman81 6 · 0 0

(Please ignore periods.. they are to displace the calculations accordingly)....

Reversed Measures: Quantum Realty and Proposed Theory:

... 2x + 2...... 4x
... --------..... ------
.... x + 1...=....1x
.... ------- ..... ------
....... x............. x

Checked Answer:

.. x = 1 * 1x = 1x^2

...... 4x
.... -------- = 4x * 1x^2 = 4x^2
..... 1x^2

True Answer & work to show:

Of 4x * 1x^2 = 4x^2 =

........ 2 ( + 1 )................... ( 2 + 1 )
. [. ----------------. + 1. ] =. --------------
......... x + 1............................ x

... 2 ( x + 1 )....... + 1..... 2 + 1........ 3
.. --------------. +. ------ = --------- = ------
..... ( x + 1)........... x........ x............ x

Answer is:

......... 3
.... -----------
......... x

Hope this helped.... Good Day

--Rob :)~

2006-07-25 18:48:11 · answer #6 · answered by stealth_n700ms 4 · 0 0

try to write more mathematically in future.

2006-07-25 20:22:11 · answer #7 · answered by rajesh bhowmick 2 · 0 0

You REALLY need to get laid.

- Regards,
Ward

2006-07-25 18:36:26 · answer #8 · answered by Anonymous · 0 0

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