Question Number 159034 by HongKing last updated on 12/Nov/21
Answered by mindispower last updated on 12/Nov/21
$${a}=\sqrt[{{n}}]{\frac{\mathrm{11}+{x}}{\mathrm{11}−{x}}},{b}=\sqrt{\frac{\mathrm{2021}+{x}}{\mathrm{2021}−{x}}} \\ $$$${a}+\frac{\mathrm{1}}{{a}}={b}+\frac{\mathrm{1}}{{b}} \\ $$$$\left({a}−{b}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{ab}}\right)=\mathrm{0} \\ $$$${a}={b},{ab}=\mathrm{1} \\ $$$${a}={b}\Rightarrow\frac{\mathrm{11}+{x}}{\mathrm{11}−{x}}=\frac{\mathrm{2021}+{x}}{\mathrm{2021}−{x}}\:,{x}=\mathrm{0} \\ $$$${ab}=\mathrm{1}\Rightarrow\left(\mathrm{11}+{x}\right)\left(\mathrm{2021}−{x}\right)=\left(\mathrm{2021}+{x}\right)\left(\mathrm{11}−{x}\right) \\ $$$$\Rightarrow\mathrm{2010}{x}=−\mathrm{2010}{x}\Rightarrow{x}=\mathrm{0} \\ $$$${S}=\left\{\mathrm{0}\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by HongKing last updated on 12/Nov/21
$$\mathrm{Cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$