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Question-140534




Question Number 140534 by SOMEDAVONG last updated on 09/May/21
Answered by EDWIN88 last updated on 09/May/21
(i) = ((16x−24)/((x−1)(x−3)(x+3))) = (a/(x−1))+(b/(x−3))+(c/(x+3))  a = [((16x−24)/((x−3)(x+3))) ]_(x=1) = ((−8)/(−8)) =1  b= [((16x−24)/((x−1)(x+3))) ]_(x=3) = ((48−24)/(6.2))=2  c = [((16x−24)/((x−1)(x−3))) ]_(x=−3) =((−72)/(−4.(−6)))=−3  ((16x−24)/((x−1)(x^2 −9))) = (1/(x−1))+(2/(x−3))−(3/(x+3))
$$\left(\mathrm{i}\right)\:=\:\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:=\:\frac{\mathrm{a}}{\mathrm{x}−\mathrm{1}}+\frac{\mathrm{b}}{\mathrm{x}−\mathrm{3}}+\frac{\mathrm{c}}{\mathrm{x}+\mathrm{3}} \\ $$$$\mathrm{a}\:=\:\left[\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:\right]_{\mathrm{x}=\mathrm{1}} =\:\frac{−\mathrm{8}}{−\mathrm{8}}\:=\mathrm{1} \\ $$$$\mathrm{b}=\:\left[\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:\right]_{\mathrm{x}=\mathrm{3}} =\:\frac{\mathrm{48}−\mathrm{24}}{\mathrm{6}.\mathrm{2}}=\mathrm{2} \\ $$$$\mathrm{c}\:=\:\left[\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right)}\:\right]_{\mathrm{x}=−\mathrm{3}} =\frac{−\mathrm{72}}{−\mathrm{4}.\left(−\mathrm{6}\right)}=−\mathrm{3} \\ $$$$\frac{\mathrm{16x}−\mathrm{24}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{x}−\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{x}+\mathrm{3}} \\ $$
Commented by SOMEDAVONG last updated on 09/May/21
  Thanks so much
$$ \\ $$$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}\: \\ $$

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