Question

If it isn't too much trouble, can someone solve and show work?

Answer 1

Factorize all things first

Answer 2

\[(x + 1)(x + 1)\]

Answer 3

\[x(1-2)\]

Answer 4

\[(x- 1)(x + 1)\]

Answer 5

\[\Large\frac{\frac{x^2+2x+1}{x-2}}{\frac{x^2-1}{x^2-4}}=\frac{\frac{(x+1)(x+1)}{(x-2)}}{\frac{(x-1)(x+1)}{(x-2)(x+2)}}=\frac{(x+1)(x+1)}{x-2}\times\frac{(x-2)(x+2)}{(x-1)(x+2)}\]

Answer 6

\[(x - 2)(x + 2)\]

Answer 7

\[\Large=\frac{(x+1)(x+1)\color{red}{(x-2)}\color{green}{(x+2)}}{\color{red}{(x-2)}(x-1)\color{green}{(x+2)}}=\frac{(x+1)(x+1)}{x-1}\]

Answer 8

Learn \(\LaTeX\), it's so fun and useful

Answer 9

That's not an answer choice D:

Answer 10

sorry i did a careless mistake

Answer 11

but I don't have mood to correct it :/

Just factorize all the stuff, and use the identity \(\Huge\frac{\left(\frac ab\right)}{\left(\frac cd\right)}=\frac ab\times\frac dc\)

Answer 12

can you please try? :D

Answer 13

\[\frac{\frac{(x+1)(x+1)}{(x-2)}}{\frac{(x-1)(x+1)}{(x-2)(x+2)}}=\frac{(x+1)(x+1)}{(x-2)}\times\frac{(x-2)(x+2)}{(x-1)(x+1)}=\frac{(x+1)\color{green}{(x+1)}\color{red}{(x-2)}(x+2)}{\color{red}{(x-2)}(x-1)\color{green}{(x+1)}}\]

Answer 14

Do the last step yourself: cancellation :)

Answer 15

thanks

Answer 16

no problem :)