Question

Find this one.... {[1+(x-1)/2]/1-1/x}-x/(x-1)
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x/2
I really hope u'll get it

Answer 1

Thank you for the star.

Answer 2

Uw :)...any idea abt this 1 ?

Answer 3

Let me give it a try

Answer 4

Thnx ^_^ :)

Answer 5

\[\frac{x-2}{x-1}\]

Answer 6

@dumbcow Will you please explain me how did u get that? Please...

Answer 7

just do it step by step, combine fractions and use property of dividing fractions (Flip and multiply)

Answer 8

\[\frac{((1+(x-1)/2)/1-1/x)-x/(x-1)}{\frac{x}{2}}=\frac{((-2+x) (-1+x))}{ x^2} \]

Answer 9

thnx :)

Answer 10

In a rush to get out a solution I did not verify the result. ie: x=17 shows that the left side and the right side are not equal.

Recalculated and it appears that at first glance that the fraction is equal to one.

Answer 11

it's ok :) I'll check it tomorrow :):):) Thnx anyway

Answer 12

dumbcow can u help me

Answer 13

The fraction is equal to 1.
The problem fraction with the braces and brackets changed to parentheses:\[\frac{((1 + (x - 1)/2)/1 - 1/x) - x/(x - 1)}{\frac{x}{2}} \]Multiply the numerator by the reciprocal of the denominator adding surrounding parentheses to the numerator.\[\frac{2}{x}(((1 + (x - 1)/2)/1 - 1/x) - x/(x - 1)) \]The following is a step by step sequence of fraction modifications, mainly simplifications to individual terms.\[\frac{2}{x}\left(((1+(x-1)/2)/1-1/x)-\frac{x}{-1+x}\right) \]\[\frac{2}{x}\left(\left(\left.\frac{1+x}{2}\right/1-1/x\right)-\frac{x}{-1+x}\right) \]\[\frac{2}{x}\left(\left(\frac{1+x}{2}/\frac{-1+x}{x}\right)-\frac{x}{-1+x}\right) \]\[\frac{2}{x}\left(\frac{x (1+x)}{2 (-1+x)}-\frac{x}{-1+x}\right) \]Multiply through by 2/x\[\left(\frac{2}{x}\right)\left(\frac{x (1+x)}{2 (-1+x)}\right)-\left(\frac{x}{-1+x}\right)\left(\frac{2}{x}\right) \]Expand the products on each side of the minus sign and simplify.\[\frac{1}{-1+x}+\frac{x}{-1+x}-\frac{2}{-1+x} \]\[\frac{1+x-2}{-1+x} \]\[\frac{-1+x}{-1+x}=1 \]

Answer 14

Wooooow...Thanks a loooooooooooooooooooooooooooot!!!!!!!!! :) ^_^ Thanks :)