Question Number 204082 by depressiveshrek last updated on 05/Feb/24
$$\sqrt{\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }+{x}}{{x}}}−\sqrt{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }−{x}^{\mathrm{2}} }=\mathrm{5} \\ $$
Commented by Frix last updated on 05/Feb/24
$${x}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{119}}} \\ $$
Commented by depressiveshrek last updated on 05/Feb/24
$${Where}\:{is}\:{the}\:{solution}? \\ $$
Commented by Frix last updated on 06/Feb/24
Inside my �� but soon out here ��
Answered by Frix last updated on 06/Feb/24
$$\mathrm{Obviously}\:{x}>\mathrm{0} \\ $$$$\mathrm{Let}\:{t}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }}{\mathrm{66}}\:\Leftrightarrow\:{x}=\frac{\mathrm{33}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}}\:\Rightarrow\:{t}>\mathrm{1} \\ $$$$\frac{\sqrt{\mathrm{2}}{t}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}−\frac{\mathrm{33}\sqrt{\mathrm{2}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}}{{t}}=\mathrm{5} \\ $$$$\mathrm{Let}\:{u}={t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\:\Leftrightarrow\:{t}=\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{u}}\:\Rightarrow\:{u}>\mathrm{1} \\ $$$$\frac{\mathrm{66}\sqrt{\mathrm{2}}}{{u}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{{u}^{\mathrm{2}} −\mathrm{1}}−\mathrm{32}\sqrt{\mathrm{2}}=\mathrm{5} \\ $$$$\mathrm{Let}\:{v}={u}^{\mathrm{2}} \:\Leftrightarrow\:{u}=\sqrt{{v}}\:\Rightarrow\:{v}>\mathrm{1}\:\mathrm{and}\:\mathrm{transforming} \\ $$$${v}^{\mathrm{2}} −\frac{\mathrm{4}\left(\mathrm{64}−\mathrm{5}\sqrt{\mathrm{2}}\right)}{\mathrm{119}}+\frac{\mathrm{2073}−\mathrm{320}\sqrt{\mathrm{2}}}{\mathrm{2023}}=\mathrm{0} \\ $$$${v}=\frac{\mathrm{123}+\mathrm{22}\sqrt{\mathrm{2}}}{\mathrm{119}}\:\Rightarrow\:{u}=\frac{\mathrm{11}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{119}}}\:\Rightarrow\:{t}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{119}}}\:\Rightarrow \\ $$$${x}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{119}}} \\ $$