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Question-140529




Question Number 140529 by SOMEDAVONG last updated on 09/May/21
Answered by Rasheed.Sindhi last updated on 09/May/21
(1/(x^2 (x+2)))−(5/((x−2)(x+2)))−(4/(x−2))  =((1(x−2)−5(x^2 )−4( x^2 (x+2) ))/(x^2 (x−2)(x+2)))  =((x−2−5x^2 −4 x^3 −8x^2  )/(x^2 (x−2)(x+2)))  =((−4x^3 −13x^2 +x−2 )/(x^2 (x−2)(x+2)))
$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{2}\right)}−\frac{\mathrm{5}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)}−\frac{\mathrm{4}}{{x}−\mathrm{2}} \\ $$$$=\frac{\mathrm{1}\left({x}−\mathrm{2}\right)−\mathrm{5}\left({x}^{\mathrm{2}} \right)−\mathrm{4}\left(\:{x}^{\mathrm{2}} \left({x}+\mathrm{2}\right)\:\right)}{{x}^{\mathrm{2}} \left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$$=\frac{{x}−\mathrm{2}−\mathrm{5}{x}^{\mathrm{2}} −\mathrm{4}\:{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} \:}{{x}^{\mathrm{2}} \left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$$=\frac{−\mathrm{4}{x}^{\mathrm{3}} −\mathrm{13}{x}^{\mathrm{2}} +{x}−\mathrm{2}\:}{{x}^{\mathrm{2}} \left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$

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