Question Number 187874 by mustafazaheen last updated on 23/Feb/23
$${how}\:{is}\:{solution} \\ $$$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{13}} =\mathrm{x}\:\:\:\:\:\:\:\:\:\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{221}} =? \\ $$$$\left.\mathrm{1}\left.\right)\left.\mathrm{x}^{−\mathrm{16}} \left.\:\:\:\:\:\:\:\:\:\:\mathrm{2}\right)\mathrm{x}^{−\mathrm{17}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{x}^{\mathrm{221}} \:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\right)\mathrm{x}^{\mathrm{21}} \\ $$
Answered by som(math1967) last updated on 23/Feb/23
$$\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{13}} ={x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{13}} }={x} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{13}} =\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow\left\{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{13}} \right\}^{\mathrm{17}} =\frac{\mathrm{1}}{{x}^{\mathrm{17}} } \\ $$$$\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{221}} ={x}^{−\mathrm{17}} \\ $$
Answered by Sutrisno last updated on 23/Feb/23
$$\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{221}} .\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{221}} }{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{221}} } \\ $$$$=\frac{\mathrm{1}}{\left(\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{13}} \right)^{\mathrm{17}} } \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{17}} } \\ $$$$={x}^{−\mathrm{17}} \\ $$