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jamesk486
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x is not 1, (1-x^5)/(1-x) < 1/1-x ?
(1) x>0
(2) x<1
using (1-x^5)/(1-x) < 1/1-x)
we can arrange the equation to be (1/1-x)- (x^5/1-x) <1/1-x
then rearrange again and get -x^5/1-x <0
multiplying each side by (1-x)^2 since its positive
=> -x^5(1-x) which is also x^5*(x-1)<0
dividing each side by x^4, we get x(x-1)<0, or 0<x<1 so the answer must be c
but what if from -x^5/(1-x)<0, instead of making it into x^5*(x-1)<0,
can u just take -x^5/(1-x)<0 and switch it to x^5/(1-x)>0 (taking out the negative sign?
for some reason if i do it that way my answer comes out differently..
(the answer is C btw)
-x^5/(1-x)<0, so is it legal to do this?
==> x^5/(1-x)>0
so is there a difference between x^5/(1-x)>0
and x^5*(x-1)<0 ?
my answer comes out different when i multiply each side by a negative (x^5/(1-x)>0 )
(1) x>0
(2) x<1
using (1-x^5)/(1-x) < 1/1-x)
we can arrange the equation to be (1/1-x)- (x^5/1-x) <1/1-x
then rearrange again and get -x^5/1-x <0
multiplying each side by (1-x)^2 since its positive
=> -x^5(1-x) which is also x^5*(x-1)<0
dividing each side by x^4, we get x(x-1)<0, or 0<x<1 so the answer must be c
but what if from -x^5/(1-x)<0, instead of making it into x^5*(x-1)<0,
can u just take -x^5/(1-x)<0 and switch it to x^5/(1-x)>0 (taking out the negative sign?
for some reason if i do it that way my answer comes out differently..
(the answer is C btw)
-x^5/(1-x)<0, so is it legal to do this?
==> x^5/(1-x)>0
so is there a difference between x^5/(1-x)>0
and x^5*(x-1)<0 ?
my answer comes out different when i multiply each side by a negative (x^5/(1-x)>0 )