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 December 24th, 2017, 02:51 PM #1 Newbie   Joined: Jan 2014 Posts: 17 Thanks: 0 Equality of two expressions Are the following two expressions equal? $\displaystyle (-4(x^5-2))/$$\displaystyle ((8+x^5)^2\sqrt{\frac{x}{5x^5+40}}) \displaystyle (4\sqrt{5}(2-x^5))/$$\displaystyle \sqrt{x(8+x^5)^3}$
 December 24th, 2017, 03:36 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,527 Thanks: 1750 For real values of x such that no denominator is zero, yes.
December 24th, 2017, 08:17 PM   #3
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Quote:
 Originally Posted by woo Are the following two expressions equal? $\displaystyle (-4(x^5-2))/$$\displaystyle ((8+x^5)^2\sqrt{\frac{x}{5x^5+40}}) \displaystyle (4\sqrt{5}(2-x^5))/$$\displaystyle \sqrt{x(8+x^5)^3}$
Yes, it is straight forward once you see that $5x^5 + 40 = 40 + 5x^5 = 5(8 + x^5).$

$\dfrac{-\ 4(x^5 - 2)}{(8 + x^5)^2 * \sqrt{\dfrac{x}{5x^5 + 40}}} = \dfrac{4(2 - x^5)}{(8 + x^5)^2 * \sqrt{\dfrac{x}{5x^5 + 40}}} = \dfrac{4(2 - x^5)}{ \sqrt{(8 + x^5 )^4} * \sqrt{\dfrac{x}{5x^5 + 40}}} =$

$\dfrac{4(2 - x^5)}{\sqrt{(8 + x^5)^4} * \sqrt{\dfrac{x}{5(8 + x^5)}}} = \dfrac{4(2 - x^5)}{\sqrt{\dfrac{x(8 + x^5)^4}{5(8 + x^5)}}} = \dfrac{4(2 - x^5)}{\sqrt{\dfrac{x(8 + x^5)^3}{5}}} =$

$\dfrac{4(2 - x^5)}{\dfrac{\sqrt{x(8 + x^5)^3}}{\sqrt{5}}} = \dfrac{4 \sqrt{5} * (2 - x^5)}{\sqrt{x(8 + x^5)^3}}.$

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