(2/x)+(2/x-1)-(2/x-2) explain how to simplify and list restrictions.
\[\large \frac{2}{x}+\frac{2}{x-1}-\frac{2}{x-2}\]
from here we can do the restrictions first...
\[\large x \neq0,1,2\]
correct?
the common denominator for all terms is x(x-1)(x-2) right?
ok... thanks, btw... so to get that common denominator in that first term we must multiply (x-1)(x-2) to the numerator and denominator correct?
yes
so in the numerator (of that first term), you'll get 2(x-1)(x-2) now for the second term, we must multiply x(x-2) to both numerator and denominator... correct?
right
what do you think we're gonna multiply to that third term?
x
remember, we're trying to get all these fraction a common denominator. that last one already has a (x-2) in its denominator. so we must multiply x(x-1)
oh k so u just have to get a common denominator right?
yes... that's what we've been doing.. but in order to get the common denominator for each fraction, we must multiply whatever we needed in to bottom and multiply that same thing to the numerator...
still with me?
yea
so then the denominator will be x(x-2)(x-1)? right?
very good!
and because we now have a common denominator, all we need to do is combine the numerators...
how
\[\large \frac {2(x-1)(x-2)+2(x)(x-2)-2(x)(x-1)}{x(x-1)(x-2)}\]
yup... you'll have to multiply that out... it's not that bad...
wao ok so it'll be 2(x^2-4x+2)?
lemme check...
yeah i have a pretty fancy calculator lol
yeah... me too... what u got?
calculator type...
2(x^2-4x+2)
yeah... that's what my calculator gave me too. i'm asking what kind of calculator u got?
oh wolfram
same...
so here's your answer... wait..
\[\large \frac {2(x^2-4x+2)}{x(x-1)(x-2)}\] you can ask wolfram to multiply out that denominator if you want... :)
hold up can u help me with some more complicated ones?
post em up as a new question.. i can't scroll to much... getting dizzy..
k
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