Question

1/9-1/x over 1/81-1/x^2

Answer 1

??

Answer 2

1 over 9 subtract 1 over x those two factors are then over 1 over 81 subtract 1 over x^2

Answer 3

multiply everything by x^2
x^2/9 -x = x^2/81 -1
x^2/9 - x^2/81 - x = -1

729x^2/81 - x^2/81 - x = 81
728x^2/81 - x = 81

this is seriously complicated-___-

Answer 4

i know right

Answer 5

sorry

Answer 6

sory that's 729x^2/81 -x = -1....not 81...maybe you'll have an idea

Answer 7

answer choices are : x+9 over 9 or 9 over x+9 or x+9 over 9x or 9x over x+9

Answer 8

please help

Answer 9

nevermind i got it

Answer 10

1/9-1/x over 1/81-1/x^2
\[\huge\frac{\frac{1}{9}-\frac{1}{x}}{\frac{1}{81}-\frac{1}{x^2}}\]

Answer 11

is this your problem?

Answer 12

oh you got it

Answer 13

yea thanks can u help on my next question

Answer 14

i suppose, post it

Answer 15

1/x-3+4/x/4/x-1/x-3

Answer 16

you're going to have to use parentheses so i know what you are asking...

Answer 17

1 over x-3 + 4 over x over 4/x - 1/x-3

Answer 18

\[\huge\frac{\frac{1}{x-3}+\frac{4}{x}}{\frac{4}{x}-\frac{1}{x-3}}\]

Answer 19

yes

Answer 20

\[\huge\frac{\frac{x+4(x-3)}{x(x-3)}}{\frac{4(x-3)-x}{x(x-3)}}\]

Answer 21

\[\huge{\frac{x+4(x-3)}{x(x-3)}}*\frac{x(x-3)}{4(x-3)-x}\]

Answer 22

\[\frac{x+4(x-3)}{4(x-3)-x}\]

Answer 23

\[\frac{x+4x-12}{4x-12-x}=\frac{5x-12}{3(x-4)}\]

Answer 24

Do you understand how to do these?

Answer 25

you must multiply each term in the numerator by the LCD of both terms, same thing with denominator

Answer 26

i know how to setit up but i havent been able to find the answer

Answer 27

The lcd of both the numerator and denominator are x(x-3), they will cancel because they divide by each other

Answer 28

is the answer 5x^3/14

Answer 29

no the answer is as i wrote it in my last post

Answer 30

it can't be factored any further

Answer 31

so the answrs 1
?

Answer 32

oo nvm

Answer 33

thanks