Question Number 172027 by Mikenice last updated on 23/Jun/22
$${solve} \\ $$$$\frac{\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}=\frac{\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}} \\ $$
Commented by Mikenice last updated on 23/Jun/22
$${thanks}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Jun/22
$${Let}\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}={y} \\ $$$$\frac{\mathrm{5}}{{y}+\mathrm{1}}=\frac{\mathrm{3}}{{y}}−\frac{\mathrm{4}}{{y}+\mathrm{7}} \\ $$$$\mathrm{5}{y}\left({y}+\mathrm{7}\right)=\mathrm{3}\left({y}+\mathrm{1}\right)\left({y}+\mathrm{7}\right)−\mathrm{4}{y}\left({y}+\mathrm{1}\right) \\ $$$$\mathrm{5}{y}^{\mathrm{2}} +\mathrm{35}{y}=\mathrm{3}{y}^{\mathrm{2}} +\mathrm{24}{y}+\mathrm{21}−\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y} \\ $$$$\mathrm{6}{y}^{\mathrm{2}} +\mathrm{15}{y}−\mathrm{21}=\mathrm{0} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} +\mathrm{5}{y}−\mathrm{7}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left(\mathrm{2}{y}+\mathrm{7}\right)=\mathrm{0} \\ $$$${y}=\mathrm{1}\:\mid\:{y}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=\mathrm{1}\:\mid\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${x}\left({x}+\mathrm{4}\right)=\mathrm{0}\:\mid\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{9}=\mathrm{0} \\ $$$$\left({x}=\mathrm{0}\:\mid\:{x}=−\mathrm{4}\right)\:\mid\:{x}=\frac{−\mathrm{8}\pm\sqrt{\mathrm{64}−\mathrm{72}}}{\mathrm{4}} \\ $$$$\left({x}=\mathrm{0}\:\mid\:{x}=−\mathrm{4}\right)\:\mid\:{x}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Commented by Mikenice last updated on 23/Jun/22
$${thanks}\:{sir} \\ $$