Question Number 120229 by Algoritm last updated on 30/Oct/20 Answered by benjo_mathlover last updated on 30/Oct/20 $$\sqrt{\mathrm{36}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{x}}\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{36}\left({x}+\mathrm{1}\right)}\:=\:{x}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{36}\left(\left[{x}+\mathrm{1}\right]^{\mathrm{2}} −\mathrm{2}{x}\right)−{x}}\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{36}\left({x}+\mathrm{1}\right)}\:=\:{x}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{36}\left({x}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 120228 by mathace last updated on 30/Oct/20 $${Please}\:{solve}\:{using}\:{special}\:{function}\:{like}\:{LambertW} \\ $$$${function}\:{or}\:{any}\:{other}\:{function}\:\left({if}\:{possibe}\right) \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}^{{x}} =\mathrm{25}{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 185760 by mathlove last updated on 27/Jan/23 $${A}=\mathrm{2}×\mathrm{10}^{\mathrm{0}} +\mathrm{10}^{−\mathrm{1}} +\mathrm{6}×\mathrm{10}^{−\mathrm{2}} +\mathrm{6}×\mathrm{10}^{−\mathrm{3}} +\mathrm{6}×\mathrm{10}^{−\mathrm{4}} +…. \\ $$$$\frac{{A}}{\mathrm{13}}=? \\ $$ Answered by Rasheed.Sindhi last updated on…
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Question Number 185734 by Shrinava last updated on 26/Jan/23 $$\mathrm{x}\:=\:\sqrt{\mathrm{8}\:+\:\sqrt{\mathrm{40}\:+\:\mathrm{8}\:\sqrt{\mathrm{5}}}} \\ $$$$\mathrm{y}\:=\:\sqrt{\mathrm{8}\:−\:\sqrt{\mathrm{40}\:+\:\mathrm{8}\:\sqrt{\mathrm{5}}}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{y}^{\mathrm{6}} \:=\:? \\ $$ Answered by abdullarasool last updated on 26/Jan/23…
Question Number 120185 by mr W last updated on 29/Oct/20 $${solve}\:{for}\:{x},\:{y}\in{Z} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\sqrt{\mathrm{2020}} \\ $$ Answered by floor(10²Eta[1]) last updated on 29/Oct/20 $$\sqrt{\mathrm{y}}=\sqrt{\mathrm{2020}}−\sqrt{\mathrm{x}} \\ $$$$\left(\mathrm{y}=\mathrm{2020}+\mathrm{x}−\mathrm{2}\sqrt{\mathrm{2020x}}\right)\:\in\mathbb{Z}…
Question Number 185723 by mnjuly1970 last updated on 26/Jan/23 Answered by cortano1 last updated on 26/Jan/23 $$\:{g}'\left({x}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}\:.{f}\left(\mathrm{1}+{x}\right)−\sqrt[{\mathrm{3}}]{{x}}\:.{f}\:'\left(\mathrm{1}+{x}\right)}{\left[{f}\left(\mathrm{1}+{x}\right)\right]^{\mathrm{2}} } \\ $$$$\:{g}\:'\left(\mathrm{1}\right)=\frac{\frac{\mathrm{1}}{\mathrm{3}}.\left(−\mathrm{4}\right)−\mathrm{1}.\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}.\left(−\frac{\mathrm{8}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{6}}\right)=−\frac{\mathrm{9}}{\mathrm{16}.\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{3}}{\mathrm{32}}…
Question Number 54641 by Joel578 last updated on 08/Feb/19 $$\mathrm{Given} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{4}{x}\:+\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:−\:\mathrm{1}}}{\:\sqrt{\mathrm{2}{x}\:+\:\mathrm{1}}\:−\:\sqrt{\mathrm{2}{x}\:−\:\mathrm{1}}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$${f}\left(\mathrm{13}\right)\:+\:{f}\left(\mathrm{14}\right)\:+\:{f}\left(\mathrm{15}\right)\:+\:…\:+\:{f}\left(\mathrm{112}\right) \\ $$ Commented by Meritguide1234 last updated on…