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

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