College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.5 - Page 76: 141

Answer

$\frac{-10}{(x-5)^{3/2}(x+5)^{1/2} }$

Work Step by Step

$(x-5)^{-1/2}(x+5)^{-1/2}-(x+5)^{1/2}(x-5)^{-3/2}$ This expression can be written as $=\frac{1}{(x-5)^{1/2}(x+5)^{1/2}} - \frac{(x+5)^{1/2}}{(x-5)^{3/2}}$ Because $[a^{-m/n} =\frac{1}{a^{m/n}} ]$ Take the LCM $= \frac{(x-5)^{1}-(x+5)^{1/2}.(x+5)^{1/2}}{(x-5)^{3/2}(x+5)^{1/2}}$ $= \frac{(x-5)^{1}-(x+5)^{1/2+1/2}}{(x-5)^{3/2}(x+5)^{1/2}}$ Because $[a^{m}\times a^{n} = a^{m+n}]$ $= \frac{(x-5)-(x+5)}{(x-5)^{3/2}(x+5)^{1/2}}$ Simplify the numerator $= \frac{(x-5-x-5)}{(x-5)^{3/2}(x+5)^{1/2}}$ $= \frac{(-10)}{(x-5)^{3/2}(x+5)^{1/2}}$
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