Question Number 188753 by BaliramKumar last updated on 19/Mar/23 $$\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}\:\:\:\:\:\:\:\:\:{x},\:{y}\:\in\:\mathrm{N} \\ $$$$\:\mathrm{H}{ow}\:{many}\:{pair}\:{of}\:\left({x},\:{y}\right) \\ $$ Answered by BaliramKumar last updated on 19/Mar/23 $${solution}:− \\…
Question Number 123175 by mathocean1 last updated on 23/Nov/20 $${Show}\:{by}\:{recurrence}\:{that} \\ $$$$\forall\:{n}\:\in\:\mathbb{N},\:\sum_{{k}=\mathrm{0}\:} ^{{n}−\mathrm{1}} {q}^{{k}} =\frac{{q}^{{n}} −\mathrm{1}}{{q}−\mathrm{1}}\: \\ $$ Answered by PNL last updated on 24/Nov/20…
Question Number 57635 by Tawa1 last updated on 09/Apr/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cubes}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{even}\:\mathrm{number},\:\:\mathrm{and} \\ $$$$\mathrm{first}\:\mathrm{n}\:\mathrm{odd}\:\mathrm{number}. \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19 $${As}\:{per}\:{higher}\:{algebra}\:{Bernard}\:{and}\:{child} \\ $$$${we}\:{can}\:{find}\: \\…
Question Number 188704 by mathlove last updated on 05/Mar/23 Answered by a.lgnaoui last updated on 05/Mar/23 $${posons}\:\:\:{Z}=\:\:{avec}:{A}={x}\sqrt{{x}}\:−{x}\:\:={x}\left(\sqrt{{x}}\:\:−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:{B}=\:\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }\:−\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}\:−\mathrm{1}}\:\:\:−\sqrt{{x}}\:\:\:\:\:\:\:{C}=\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} \:}\:+\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}−\sqrt{{x}\:}\: \\ $$$${B}=\frac{\:\:\sqrt{{x}}\:−\mathrm{1}\:}{\:\sqrt[{\mathrm{4}}]{{x}}\:−\mathrm{1}}\:\:{C}=\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}\:\:{B}×{C}=\mathrm{1}−\sqrt{{x}}\: \\ $$$$\Rightarrow\:\frac{{A}}{{B}×{C}}=\frac{{x}\left(\sqrt{{x}}\:−\mathrm{1}\right)}{\:\sqrt{{x}}\:−\mathrm{1}}\:\:\:\:…
Question Number 188660 by mr W last updated on 04/Mar/23 $${solve}\:{x}^{\mathrm{4}} +\mathrm{4}{x}=\mathrm{1} \\ $$ Answered by aba last updated on 04/Mar/23 $$\bullet\mathrm{x}^{\mathrm{4}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}}…
Question Number 188653 by Shrinava last updated on 04/Mar/23 $$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{different}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{the} \\ $$$$\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word}\:\mathrm{ABRAKADABRA} \\ $$$$\mathrm{be}\:\mathrm{arranged}? \\ $$ Commented by Ajetunmobi last updated on 04/Mar/23 $$\frac{\mathrm{11}!}{\mathrm{5}!×\mathrm{2}!×\mathrm{2}!}=\mathrm{83160} \\…
Question Number 57585 by Mikael_Marshall last updated on 07/Apr/19 $${knowing}\:{that}\:{x}+{y}=\mathrm{1}.\:{what}\:{is}\:{the}\:{result}\:{of}\:\frac{{y}}{{x}}+\frac{{x}}{{y}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19 $$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{xy}} \\ $$$${R}=\frac{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}}{{xy}}…
Question Number 123114 by nimnim last updated on 23/Nov/20 $${Is}\:{there}\:{any}\:{solution}\left({s}\right)?? \\ $$$$\begin{cases}{\mathrm{36}{x}^{\mathrm{2}} {y}−\mathrm{27}{y}^{\mathrm{3}} =\mathrm{8}}\\{\mathrm{4}{x}^{\mathrm{3}} −\mathrm{27}{xy}^{\mathrm{2}} =\mathrm{4}}\end{cases} \\ $$$${please}…. \\ $$ Commented by benjo_mathlover last updated…
Question Number 188648 by Shrinava last updated on 04/Mar/23 Answered by SEKRET last updated on 04/Mar/23 $${Q}\mathrm{188572} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 188647 by Rupesh123 last updated on 04/Mar/23 Commented by mr W last updated on 05/Mar/23 $$\mathrm{4}. \\ $$$$\underset{{max}} {\mathrm{3}}^{\:\:\:\frac{\mathrm{1}}{\mathrm{3}}} >\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}}} >\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{5}}} >\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{6}}}…