Question Number 163284 by MathsFan last updated on 05/Jan/22
$$\:{given}\:\:\mathrm{2020}^{{x}} +\mathrm{2020}^{−{x}} =\mathrm{3} \\ $$$$\:\sqrt{\frac{\mathrm{2020}^{\mathrm{6}{x}} −\mathrm{2020}^{−\mathrm{6}{x}} }{\:\mathrm{2020}^{{x}} −\mathrm{2020}^{−{x}} }}=? \\ $$
Answered by Rasheed.Sindhi last updated on 05/Jan/22
$$\:\mathrm{2020}^{{x}} +\mathrm{2020}^{−{x}} =\mathrm{3} \\ $$$$\:\sqrt{\frac{\mathrm{2020}^{\mathrm{6}{x}} −\mathrm{2020}^{−\mathrm{6}{x}} }{\:\mathrm{2020}^{{x}} −\mathrm{2020}^{−{x}} }}=? \\ $$$$\mathrm{2020}^{{x}} ={y}\:,\:\mathrm{2020}^{−{x}} =\frac{\mathrm{1}}{{y}} \\ $$$$\:\:{y}+\frac{\mathrm{1}}{{y}}=\mathrm{3} \\ $$$$\:\left(\:{y}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${y}^{\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=\mathrm{9}−\mathrm{2}=\mathrm{7} \\ $$$$\:\sqrt{\frac{\mathrm{2020}^{\mathrm{6}{x}} −\mathrm{2020}^{−\mathrm{6}{x}} }{\:\mathrm{2020}^{{x}} −\mathrm{2020}^{−{x}} }}=\sqrt{\frac{{y}^{\mathrm{6}} −\frac{\mathrm{1}}{{y}^{\mathrm{6}} }}{{y}−\frac{\mathrm{1}}{{y}}}} \\ $$$$=\sqrt{\frac{\cancel{\left({y}−\frac{\mathrm{1}}{{y}}\right)}\left({y}+\frac{\mathrm{1}}{{y}}\right)\left({y}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)\left({y}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)}{\cancel{\left({y}−\frac{\mathrm{1}}{{y}}\right)}}} \\ $$$$=\sqrt{\left({y}+\frac{\mathrm{1}}{{y}}\right)\left({y}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)\left({y}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)} \\ $$$$=\sqrt{\left(\mathrm{3}\right)\left(\mathrm{7}−\mathrm{1}\right)\left(\mathrm{7}+\mathrm{1}\right)}=\sqrt{\mathrm{144}}=\mathrm{12} \\ $$
Commented by MathsFan last updated on 05/Jan/22
$${merci}\:{senior} \\ $$
Commented by peter frank last updated on 06/Jan/22
$$\mathrm{great} \\ $$
Commented by Rasheed.Sindhi last updated on 06/Jan/22
$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{Peter}\:\mathrm{Frank}! \\ $$