Question Number 208866 by Adeyemi889 last updated on 26/Jun/24
Commented by Adeyemi889 last updated on 26/Jun/24
$${partial}\:{F}\boldsymbol{{Raction}} \\ $$$$ \\ $$
Answered by Sutrisno last updated on 26/Jun/24
$$=\frac{{x}^{\mathrm{5}} +\mathrm{4}{x}^{\mathrm{3}} }{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}} \\ $$$$={x}+\mathrm{2}+\frac{\mathrm{5}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{4}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{5}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{4}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{{ax}+{b}}{{x}^{\mathrm{2}} +\mathrm{2}}+\frac{{c}}{{x}−\mathrm{1}}+\frac{{d}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{5}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{4}=\left({ax}+{b}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{c}\left({x}^{\mathrm{2}} +\mathrm{2}\right)\left({x}−\mathrm{1}\right)+{d}\left({x}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$$\bullet{x}=\mathrm{1} \\ $$$$\:\:{d}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\bullet{x}=\mathrm{0} \\ $$$$\:\:\:{b}−\mathrm{2}{c}+\mathrm{2}{d}=−\mathrm{4} \\ $$$$\bullet{x}=−\mathrm{1} \\ $$$$\:\:\:−\mathrm{4}{a}+\mathrm{4}{b}−\mathrm{6}{c}+\mathrm{3}{d}=−\mathrm{17} \\ $$$$\bullet{x}=\mathrm{2} \\ $$$$\:\:\:\mathrm{2}{a}+{b}+\mathrm{6}{c}+\mathrm{6}{d}=\mathrm{40} \\ $$$${didapat}\::{a}=\frac{\mathrm{4}}{\mathrm{9}},{b}=\frac{\mathrm{16}}{\mathrm{9}},{c}=\frac{\mathrm{41}}{\mathrm{9}} \\ $$$${fraksi}\:: \\ $$$${x}+\mathrm{2}+\frac{\mathrm{4}{x}+\mathrm{16}}{\mathrm{9}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}+\frac{\mathrm{41}}{\mathrm{9}\left({x}−\mathrm{1}\right)}+\frac{\mathrm{5}}{\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$