Question Number 213374 by hardmath last updated on 03/Nov/24
$$\mathrm{Find}:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\sqrt{\mathrm{5}}}{\left(\sqrt{\mathrm{3}}\:−\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} }\:=\:? \\ $$
Commented by Frix last updated on 03/Nov/24
$$\frac{\mathrm{29}−\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{8}−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{3}}{\mathrm{4}}} +\frac{\mathrm{13}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{\mathrm{18}−\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$
Commented by hardmath last updated on 03/Nov/24
$$\mathrm{Answer}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\:\:,\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:,\:\:\:\frac{\mathrm{2}}{\mathrm{3}}\:\:\:,\:\:\:\frac{\mathrm{3}}{\mathrm{4}}\:\:\:,\:\:\:\frac{\mathrm{7}}{\mathrm{4}} \\ $$
Commented by Frix last updated on 03/Nov/24
$$\mathrm{Either}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{or}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{answers}\:\mathrm{are}\:\mathrm{wrong}. \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{expression}\:\mathrm{is} \\ $$$$\approx\mathrm{4}.\mathrm{37943662345} \\ $$
Commented by hardmath last updated on 03/Nov/24
$$\mathrm{can}\:\mathrm{be},\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by MrGaster last updated on 03/Nov/24
$$\frac{\mathrm{1}−\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{5}}+\left(\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} }{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\sqrt[{\mathrm{4}}]{\mathrm{5}}+\left(\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt{\mathrm{5}}} \\ $$$$=\left(\frac{\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\:\sqrt{\mathrm{3}}−\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right) \\ $$$$=\left(\frac{\sqrt{\mathrm{3}}+\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\:\sqrt{\mathrm{3}}+\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right) \\ $$$$=\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1} \\ $$$${ps}:\mathrm{In}\:\mathrm{the}\:\mathrm{above}\:\mathrm{equation}\:\mathrm{since}\:\mathrm{ther} \\ $$$$\mathrm{numerato}\:\mathrm{and}\:\mathrm{denominator}\:\mathrm{are}\:\mathrm{in}\:\mathrm{thee} \\ $$$$\mathrm{sam}\:\mathrm{form}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{is}\:\mathrm{actually}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1}. \\ $$$$\mathrm{The}\:\mathrm{approximate}\:\mathrm{value}\:\mathrm{is}: \\ $$$$\approx\mathrm{4}.\mathrm{3794366234524650772560550532914496820}… \\ $$
Commented by Frix last updated on 04/Nov/24
$$\mathrm{Again}\:\mathrm{nonsense} \\ $$$$\mathrm{obviously}\:\frac{\left(\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt{\mathrm{5}}}\neq\left(\frac{\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\:\sqrt{\mathrm{3}}−\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right) \\ $$$$\mathrm{obviously}\:\left(\frac{\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\:\sqrt{\mathrm{3}}−\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right)\neq\left(\frac{\sqrt{\mathrm{3}}+\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\:\sqrt{\mathrm{3}}+\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right) \\ $$$$\mathrm{obviously}\:\mathrm{1}\neq\mathrm{4}.\mathrm{379}… \\ $$
Commented by issac last updated on 04/Nov/24
$$\mathrm{lol} \\ $$
Answered by MathematicalUser2357 last updated on 06/Nov/24
$${use}\:{the}\:{editor}\:{tinku}\:{tara} \\ $$$$\frac{\mathrm{1}−\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{5}}+\sqrt{\mathrm{5}}}{\left(\sqrt{\mathrm{3}}−\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}.\mathrm{379437} \\ $$