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1 - x --- x + 1 --------- 2 - x - 1 ----- x
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\[\frac{1-\frac{x}{x+1}}{2-\frac{x-1}{x}}\]
multiply top and bottom by \[x(x+1)\] to clear the factions
that is correct
so numerator will be \[(1-\frac{x}{x+1})\times x(x+1)\] \[=x(x+1)-x^2\] denominator will be \[(2-\frac{x-1}{x})\times x(x+1)\] \[2x(x+1)-(x-1)(x+1)\] giving \[\frac{x(x+1)-x^2}{2x(x+1)-(x-1)(x+1)}\]
then multiply out and combine like terms in numerator and denominator
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numerator is \[x\] denominator is \[2x^2+2x-(x^2-1)=x^2+2x+1\] if my algebra is correct.
looks good
why thank you!
dont let it go to your head
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