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5x-2-15x-11-x-1-x-2-2-fractio-partil-




Question Number 159433 by joki last updated on 17/Nov/21
((5x^2 −15x−11)/((x+1)(x−2)^2 )) fractio partil
$$\frac{\mathrm{5x}^{\mathrm{2}} −\mathrm{15x}−\mathrm{11}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} }\:\mathrm{fractio}\:\mathrm{partil} \\ $$$$ \\ $$
Commented by cortano last updated on 17/Nov/21
 ((5x^2 −15x−11)/((x+1)(x−2)^2 )) = (a/(x+1)) +(b/(x−2)) +(c/((x−2)^2 ))   5x^2 −15x−11=a(x−2)^2 +b(x+1)(x−2)+c(x+1)  x=−1⇒9=9a ;a=1  x=2⇒−21=3c ; c=−7  x=0⇒−11=4−2b−7  ⇒−8=−2b ; b=4  ⇔ ((5x^2 −15x−11)/((x+1)(x−2)^2 ))=(1/(x+1))+(4/(x−2))−(7/((x−2)^2 ))
$$\:\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{15}{x}−\mathrm{11}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:=\:\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{b}}{{x}−\mathrm{2}}\:+\frac{{c}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\:\mathrm{5}{x}^{\mathrm{2}} −\mathrm{15}{x}−\mathrm{11}={a}\left({x}−\mathrm{2}\right)^{\mathrm{2}} +{b}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)+{c}\left({x}+\mathrm{1}\right) \\ $$$${x}=−\mathrm{1}\Rightarrow\mathrm{9}=\mathrm{9}{a}\:;{a}=\mathrm{1} \\ $$$${x}=\mathrm{2}\Rightarrow−\mathrm{21}=\mathrm{3}{c}\:;\:{c}=−\mathrm{7} \\ $$$${x}=\mathrm{0}\Rightarrow−\mathrm{11}=\mathrm{4}−\mathrm{2}{b}−\mathrm{7} \\ $$$$\Rightarrow−\mathrm{8}=−\mathrm{2}{b}\:;\:{b}=\mathrm{4} \\ $$$$\Leftrightarrow\:\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{15}{x}−\mathrm{11}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{4}}{{x}−\mathrm{2}}−\frac{\mathrm{7}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$
Commented by cortano last updated on 17/Nov/21
oo yes. i′m sorry sir
$${oo}\:{yes}.\:{i}'{m}\:{sorry}\:{sir} \\ $$

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