Question

Simplify completely:

Answer 1

hi again

Answer 2

\(\Large 9(\frac{t-2}{2t+1})^{8}\times \frac{(2t-1)-2(t-2)}{(2t+1)^{2}}\)

Answer 3

i think u know me as GTA_Hunter35 that account kinda crashed

Answer 4

^^There's the thing I have to simplify. Not quite sure where to begin.

Answer 5

@please_help_me if you'd like to address something to me, please contact me via pm. Do not spam my post, thank you :)

Answer 6

Parentheses
Exponents
Multiplication
Division
Addition
Subtraction

FROM LEFT TO RIGHT

Answer 7

all we can do is:
\[\Large \begin{gathered}
9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\
\hfill \\
= 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{3}{{{{(2t + 1)}^2}}} \hfill \\
\end{gathered} \]

Answer 8

Is that all we can do?

Answer 9

therefore, we can write this:
\[\Large \begin{gathered}
9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\
\hfill \\
= 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{3}{{{{(2t + 1)}^2}}} = \hfill \\
\hfill \\
= 27\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\
\end{gathered} \]

Answer 10

yes! I think so!

Answer 11

what about squaring now?

Answer 12

better is if we don't develop the powers of binomials, because we have to do a very long computation @please_help_me

Answer 13

well.... ur right ithink thats it

Answer 14

wait. For the top part,

(2t+1)-2(t-2) = 2t-2t+1+4? so then we would have


\(\Large \begin{gathered}
9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t - 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\
\hfill \\
= 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{\color{red}{5}}{{{{(2t + 1)}^2}}} \hfill \\
\end{gathered}\)

Answer 15

basically, a 5 instead of a 3?

Answer 16

I got
2t-1-2t+4

Answer 17

which equals 3

Answer 18

distribute, then ull get 3

Answer 19

-1? o-o

oh. wait, i think i see something.

\(\Large \begin{gathered}
9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t \color{red}+ 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\
\hfill \\
= 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{\color{red}{5}}{{{{(2t + 1)}^2}}} \hfill \\
\end{gathered}\)

Answer 20

in the original problem, it is 2t+1, not 2t-1.

Answer 21

oh wow. I wrote it out wrong. So sorry XD

Answer 22

np was a typo

Answer 23

ok! then you are right, it is 5

Answer 24

here are the updated steps:
\[\Large \begin{gathered}
9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t + 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\
\hfill \\
= 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{5}{{{{(2t + 1)}^2}}} = \hfill \\
\hfill \\
= 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\
\end{gathered} \]

Answer 25

to clarify, let me type this out correctly.
*cracks knuckles* \(\LaTeX\) is hard

\(\Large 9(\frac{t-2}{2t+1})^{8} \times \frac{(2t+1)-2(t-2)}{(2t+1)^{2}}\)

Answer 26

can u guys help me

Answer 27

So I don't have to factor out the denominator or do anything fancy like that?

Answer 28

i dont think so

Answer 29

no, since numerator and denominator don't contain common divisors

Answer 30

oh, and how did you get from the penultimate step to the last step? Like what are the exact operations that are involved in combining this multiplication problem into one fraction?

Answer 31

and yeah you're right, I don't see any common divisors. That makes sense

Answer 32

see the 2t-1?

Answer 33

I have applied the rule of multiplication of powers with same basis at denominator

Answer 34

2t-1 was to the 8th power and the other 2t-1 was to the2nd power

Answer 35

multiplying numbers with exponents adds up the exponents

Answer 36

here are more steps:
\[\large \begin{gathered}
9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{{(2t + 1) - 2(t - 2)}}{{{{(2t + 1)}^2}}} = \hfill \\
\hfill \\
= 9{\left( {\frac{{t - 2}}{{2t + 1}}} \right)^8} \times \frac{5}{{{{(2t + 1)}^2}}} = 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^8}}} \cdot \frac{1}{{{{\left( {2t + 1} \right)}^2}}} \hfill \\
\hfill \\
= 45\frac{{{{\left( {t - 2} \right)}^8}}}{{{{\left( {2t + 1} \right)}^{10}}}} \hfill \\
\end{gathered} \]

Answer 37

which created 2t-1^10

Answer 38

Oh I see! I was just kind of wondering how the numerator of five contributed to the solid number of 45 in the mixed fraction. I see it now

Answer 39

ok! :)

Answer 40

Thank you so much, it makes sense now :D

Answer 41

:)