Menu Close

Question-209129




Question Number 209129 by Adeyemi889 last updated on 02/Jul/24
Answered by A5T last updated on 02/Jul/24
((x^3 +3)/(x^2 −1))=((x(x^2 −1)+x+3)/(x^2 −1))=x+((x+3)/(x^2 −1))  ((x+3)/(x^2 −1))=(A/(x−1))+(B/(x+1))⇒x+3=(A+B)x+A−B  ⇒A+B=1;A−B=3⇒A=2;B=−1  ⇒((x^2 +3)/(x^2 −1))=x+(2/(x−1))−(1/(x+1))
$$\frac{{x}^{\mathrm{3}} +\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}={x}+\frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{{x}+\mathrm{1}}\Rightarrow{x}+\mathrm{3}=\left({A}+{B}\right){x}+{A}−{B} \\ $$$$\Rightarrow{A}+{B}=\mathrm{1};{A}−{B}=\mathrm{3}\Rightarrow{A}=\mathrm{2};{B}=−\mathrm{1} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}={x}+\frac{\mathrm{2}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *