Question Number 81403 by naka3546 last updated on 12/Feb/20
$${Solve}\:\:{x}\:\in\:\mathbb{R}\:\:\:{such}\:\:{that} \\ $$$$\frac{\left({x}+\mathrm{1}\right)^{\mathrm{5}} }{\left({x}^{\mathrm{5}} \:+\:\mathrm{1}\right)}\:\:=\:\:\frac{\mathrm{81}}{\mathrm{11}} \\ $$
Commented by MJS last updated on 12/Feb/20
$${x}\neq−\mathrm{1} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{7}}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\mathrm{2} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =−\frac{\mathrm{5}}{\mathrm{14}}\pm\frac{\mathrm{3}\sqrt{\mathrm{19}}}{\mathrm{14}}\mathrm{i} \\ $$
Answered by ajfour last updated on 12/Feb/20
$$\frac{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }=\frac{\mathrm{11}}{\mathrm{81}} \\ $$$${x}=\mathrm{2}\:. \\ $$