Question

(v-8/v-9)-(v+1/v+9)+(v-27/v^2-81)

Answer 1

-27/v^2 + v-9/v -99

Answer 2

-27/v^2 + v-(9/v) -99

Answer 3

-27/v^2 +(v^2-9)/v -99

Answer 4

there's alot of ways to writing it

Answer 5

do u want it under one fraction

Answer 6

Yes please

Answer 7

(v^3-99v^2-9v-27)/(v^2)

Answer 8

i think that's the answer

Answer 9

\[\frac{v-8}{v-9} \cdot \frac{v+9}{v+9} -\frac{v+1}{v+9} \frac{v-9}{v-9}+\frac{v-27}{(v-9)(v+9)}\]
\[=\frac{v^2-8v+9v-72-(v^2-9v+v-9)+(v-27)}{(v-9)(v+9)}\]
\[=\frac{v^2-v^2+9v-8v+9v-v+v-72+9-27}{(v-9)(v+9)}\]
\[=\frac{10v-90}{(v-9)(v+9)}\]
is what i got
but i think hahd and interpreted the problem differently

Answer 10

yes, yes i did

Answer 11

when you write (v-8/v-9) actually means to me
that you are doing \[v-\frac{8}{v}-9\]
but if you say (v-8)/(v-9) actually means to me and everyone
that you are doing \[\frac{v-8}{v-9}\]
eventhough you didn't write it this second way i mentioned i just tried to use esp

Answer 12

yh

Answer 13

oh yeah we could go alittle further on what i have

Answer 14

\[\frac{10(x-9)}{(v-9)(v+9)}=\frac{10}{v+9}, v \neq 9\]