anonymous
  • anonymous
Simplify:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1440898252857:dw|
LynFran
  • LynFran
not clear...
anonymous
  • anonymous
\[(4-x^2)(1/3)(6x+1)^{-2/3}(6)-(6x+1)^{1/3}(-2x) / (4-x^2)^2\]

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More answers

LynFran
  • LynFran
6*1/3=2 ...\[\frac{[2( 4-x ^{2})(6x+1)^{\frac{ -2 }{ 3 }}]-[(6x+1)^{\frac{ 1 }{ 3 }}(2x)] }{ (4-x ^{2})^{2} }\]
anonymous
  • anonymous
Do you start canceling?
anonymous
  • anonymous
I have the answer, and that isn't right. I need to know how to work it.
LynFran
  • LynFran
ok so we have ...
LynFran
  • LynFran
\[\frac{ 2(6x+1)^{-1/3} [(6x+1)^{-1/3}(4-x ^{2})-(x)]}{ (4-x ^{2})^{2} }\]
LynFran
  • LynFran
@Loser66 what do u think..
Nnesha
  • Nnesha
1/3 is an exponent at the numerator ?
anonymous
  • anonymous
yes
Nnesha
  • Nnesha
\[\large\rm \frac{ 6 (4-x^2)^\frac{ 1 }{ 3 } (6x+1)^\frac{ -2 }{ 3}-(6x+1)^\frac{ 1 }{ 3 }(-2x) }{ (4-x^2)^2}\]hmm like this ?
anonymous
  • anonymous
There is no 1/3 exponent for (4-x^2).
anonymous
  • anonymous
\[(4-x^2)(1/3)(6x+1)^{-2/3}-(6x+1)^{1/3}(-2x) / (4-x^2)^2\] is the problem
Nnesha
  • Nnesha
ohh i asked before
Nnesha
  • Nnesha
\[\large\rm \frac{ 6(\frac{1}{3}) (4-x^2) (6x+1)^\frac{ -2 }{ 3}-(6x+1)^\frac{ 1 }{ 3 }(-2x) }{ (4-x^2)^2}\] looks right ?
anonymous
  • anonymous
No
Nnesha
  • Nnesha
no hmm can you please take a screenshot
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @ducksonquack \[(4-x^2)(1/3)(6x+1)^{-2/3}(6)-(6x+1)^{1/3}(-2x) / (4-x^2)^2\] \(\color{blue}{\text{End of Quote}}\) there is 6
anonymous
  • anonymous
Nnesha
  • Nnesha
alright thanks
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @ducksonquack I have the answer, and that isn't right. I need to know how to work it. \(\color{blue}{\text{End of Quote}}\) what's the answer jsut want to check if i'm doing this right or not>.<
anonymous
  • anonymous
\[2(5x^2+x+4) / (4-x^2)^2(6x+?)^{2/3}\]
anonymous
  • anonymous
Didn't catch what the ? mark was when we were given the answers.
Jhannybean
  • Jhannybean
Well that's in your problem, it states \((6x+1)\) in parenthesis.... so im sure the \(? = 1\)
anonymous
  • anonymous
Okay. What's next?
Jhannybean
  • Jhannybean
One second im solving it
Jhannybean
  • Jhannybean
Alright got it, so...
anonymous
  • anonymous
?
Jhannybean
  • Jhannybean
\[=\frac{\frac{1}{3}\cdot 6 \cdot (4-x^2)(6x+1)^{-2/3}-(6x+1)^{1/3}(-2x)}{(4-x^2)^2}\]\[=\frac{(6x+1)^{-2/3}\left(2\left[4-x^2-(6x+1)(-x)\right]\right)}{(4-x^2)^2}\]\[=\frac{2\left[4-x^2+6x+x\right]}{(4-x^2)^2(6x+1)^{2/3}}\]\[=\boxed{\frac{2(5x^2+x+4)}{(4-x^2)^2(6x+1)^{2/3}}}\]
anonymous
  • anonymous
Can you explain it to me step by step?
Jhannybean
  • Jhannybean
Okay so what you want to do first is look for a common like term. in this case it would be \((6x+1)^{-2/3}\). Notice how its a common multiplier in the numerator? We can pull it out. Now I pulled out \((6x+1)^{-2/3}\) instead of the \((6x+1)^{1/3}\) because outof the two, \((6x+1)^{-2/3}\) is the smaller power, therefore pulling it out as a common factor, I am left with \((6x+1)^1\) inside the brackets. (Remember: \((6x+1)^{-2/3} \cdot (6x+1)^1 = (6x+1)^{1/3}\)) Now I pulled out a 2 inside the brackets because that is another common like term between the entire function within the brackets. \([2(4-x^2)-(6x+1)(-2x)] \iff 2[(4-x^2-(6x+1)(-x)]\) Then I simplify my fraction even further by turning \((6x+1)^{-2/3}\) positive by putting it in the denominator. this turns it positive. Recall that: \(x^{-\#} \iff \dfrac{1}{x^{\#}}\) So far we have: \[\frac{2[4-x^2-(6x+1)(-x)]}{(4-x^2)^2(6x+1)^{2/3}}\] Now we simplify the stuff inside the brackets without multiplying in the 2. \[\frac{2[4-x^2 +6x+x]}{(4-x^2)^2(6x+1)^{2/3}}\] Simplify this a bit more and you'll have: \(\boxed{ \dfrac{2[5x^2+x+4]}{(4-x^2)^2(6x+1)^{2/3}}}\)
Nnesha
  • Nnesha
sorry i was afk great explanation jhanny!
Jhannybean
  • Jhannybean
Thank you :)

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