Question Number 137591 by Ñï= last updated on 04/Apr/21 $${a}=\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{1}} \\ $$$$\frac{\mathrm{3}}{{a}}+\frac{\mathrm{3}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=? \\ $$ Commented by mr W last updated on 05/Apr/21 $$=\mathrm{1}…
Question Number 137590 by Ñï= last updated on 04/Apr/21 $${x}=\mathrm{1}+\frac{\pi+\left(\pi+\mathrm{1}\right)^{\mathrm{2}} +\left(\pi+\mathrm{2}\right)^{\mathrm{3}} +\left(\pi+\mathrm{3}\right)^{\mathrm{4}} }{\mathrm{4}+\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{3}} +\mathrm{7}^{\mathrm{4}} } \\ $$$$\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}+\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}}=? \\ $$ Commented by mindispower last updated…
Question Number 137585 by bramlexs22 last updated on 04/Apr/21 $${Find}\:{the}\:{cube}\:{of}\:{the}\:{number}\: \\ $$$${N}=\:\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}\sqrt{\mathrm{7}\sqrt{\mathrm{3}…}}}}}}}} \\ $$ Answered by bemath last updated on 04/Apr/21 $${N}=\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{32}}+\frac{\mathrm{1}}{\mathrm{128}}+…} .\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{64}}+…} \\ $$$${N}=\mathrm{7}^{\frac{\mathrm{1}/\mathrm{2}}{\mathrm{1}−\mathrm{1}/\mathrm{4}}}…
Question Number 137584 by bramlexs22 last updated on 04/Apr/21 $${Given}\:\begin{cases}{{a}_{\mathrm{2}{n}} \:=\:{a}_{{n}} .{a}_{\mathrm{2}} \:+\mathrm{1}}\\{{a}_{\mathrm{2}{n}+\mathrm{1}} \:=\:{a}_{{n}} .{a}_{\mathrm{2}} \:−\mathrm{2}\:}\end{cases}\:{and} \\ $$$$\:\begin{cases}{{a}_{\mathrm{7}} \:=\:\mathrm{2}}\\{\mathrm{0}<{a}_{\mathrm{1}} <\mathrm{1}}\end{cases}.\:{Find}\:{a}_{\mathrm{25}} \:=? \\ $$$$ \\ $$…
Question Number 72048 by ahmadshahhimat775@gmail.com last updated on 23/Oct/19 Answered by mind is power last updated on 23/Oct/19 $$\mathrm{a}_{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{1}+\mathrm{3}}=\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}.\mathrm{2}^{\mathrm{0}} }{\mathrm{2}}=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}+\mathrm{2}^{\mathrm{0}} }{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}}=\frac{\mathrm{6}}{\mathrm{5}}\:=\frac{\mathrm{3}.\mathrm{2}}{\mathrm{3}.\mathrm{2}−\mathrm{1}}=\frac{\mathrm{4}+\mathrm{2}}{\mathrm{4}+\mathrm{2}−\mathrm{1}}=\frac{\mathrm{2}+\mathrm{2}^{\mathrm{2}}…
Question Number 6512 by Rasheed Soomro last updated on 30/Jun/16 $${Find}\:{complex}\:{number}\:{whose}\:{additive}\: \\ $$$${inverse}\:{is}\:{equal}\:{to}\:{its}\:{multiplicative} \\ $$$${inverse}. \\ $$ Commented by Temp last updated on 30/Jun/16 $$\mathrm{What}\:\mathrm{is}\:\mathrm{additive}\:\mathrm{and}\:\mathrm{multiplicitive}…
Question Number 72046 by aliesam last updated on 23/Oct/19 Commented by mathmax by abdo last updated on 26/Oct/19 $${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {nln}\left(\mathrm{1}+\left(\frac{{x}}{{n}}\right)^{\alpha} \right){dx}\:\Rightarrow{U}_{{n}} =_{\frac{{x}}{{n}}={t}} \:\:\:\int_{\mathrm{0}}…
Question Number 6511 by Yozzii last updated on 30/Jun/16 $${Find}\:{the}\:{function}\:{y}\:{satisfying} \\ $$$${y}'+\frac{{c}_{\mathrm{1}} }{{y}\left({x}−{c}_{\mathrm{2}} \right)^{{n}} }={c}_{\mathrm{3}} \\ $$$${where}\:{n}\in\left(\mathbb{Z}−\left\{\mathrm{0}\right\}\right),\:{c}_{\mathrm{1}} \:{and}\:{c}_{\mathrm{3}} \:{are}\:{nonzero} \\ $$$${constants},\:{and}\:{c}_{\mathrm{2}} \:{is}\:{constant}. \\ $$ Terms…
Question Number 72047 by aliesam last updated on 23/Oct/19 Commented by mind is power last updated on 23/Oct/19 $$\mathrm{x}_{\mathrm{1}} =\mathrm{1},\mathrm{x}_{\mathrm{2}} =\mathrm{4},\mathrm{x}_{\mathrm{3}} =\mathrm{9},\mathrm{x}_{\mathrm{4}} =\mathrm{16} \\ $$$$\forall_{\mathrm{n}}…
Question Number 72044 by 20190927 last updated on 23/Oct/19 $$\mathrm{y}=\mathrm{ln}\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{x}\right)\:\:\:\mathrm{solve}\:\mathrm{y}^{\left(\mathrm{6}\right)} \left(\mathrm{x}\right) \\ $$ Commented by mathmax by abdo last updated on 24/Oct/19 $${we}\:{have}\:{y}^{'} \left({x}\right)=\frac{\mathrm{6}{x}+\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}}…