Question Number 192612 by York12 last updated on 22/May/23 $$\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}}\:=\:\mathrm{1}\: \\ $$$${find}\:{the}\:{minimum}\:{value}\:{of}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \\ $$ Commented by Frix last updated on 23/May/23 $$\mathrm{Yes}\:\mathrm{of}\:\mathrm{course}! \\…
Question Number 127078 by amns last updated on 26/Dec/20 $$\frac{\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}}{\frac{\frac{\frac{\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\mathrm{10}}}{\mathrm{3}}}{\mathrm{2}}}\:=\:? \\ $$ Commented by hknkrc46 last updated on 26/Dec/20 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\::\:\sqrt{\mathrm{2}}\right)\::\:\left[\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\::\:\mathrm{10}\::\:\mathrm{3}\::\:\mathrm{2}\right] \\ $$$$=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\::\:\left[\frac{\sqrt{\mathrm{2}}\:+\:\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\centerdot\:\frac{\mathrm{1}}{\mathrm{10}}\:\centerdot\:\frac{\mathrm{1}}{\mathrm{3}}\:\centerdot\:\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\centerdot\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{10}}{\mathrm{1}}\:\centerdot\:\frac{\mathrm{3}}{\mathrm{1}}\:\centerdot\:\frac{\mathrm{2}}{\mathrm{1}} \\…
Question Number 192608 by York12 last updated on 22/May/23 $${let}\:{k}\:{be}\:{natural}\:{number}.\:{defined}\:{s}_{{k}} \:{as}\:{the} \\ $$$${sum}\:{of}\:{the}\:{infinite}\:{series}\:{s}_{{k}} =\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{0}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{1}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }\:+… \\ $$$${find}\:{the}\:{value}\:{of}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[\frac{{s}_{{k}} }{\mathrm{2}^{{k}−\mathrm{1}}…
Question Number 61537 by aanur last updated on 04/Jun/19 Commented by aanur last updated on 04/Jun/19 $${sir}\:{could}\:{you}\:{help}\:{me} \\ $$ Commented by math1967 last updated on…
Question Number 192604 by sciencestudentW last updated on 22/May/23 $${many}\:{people}\:{say}\:{that}\:{the}\:\mathrm{0}^{\mathrm{0}\:} {is}\:{an}\:{uninfinity} \\ $$$${ones}\:{of}\:{them}\:{say}\:{that}\:\mathrm{0}^{\mathrm{0}} \:{is}\:{infinity}\:{and}\:{equal}\:{to}\:\mathrm{1}! \\ $$$${what}\:{do}\:{you}\:{think}\:{wich}\:{ones}\:{of}\:{them} \\ $$$${say}\:{right}? \\ $$ Commented by Frix last updated…
Question Number 192606 by sciencestudentW last updated on 22/May/23 $${proof}\:{that}\:\mathrm{0}^{\mathrm{0}} =\mathrm{1}\:{without}\:{limit}! \\ $$ Commented by Frix last updated on 22/May/23 $$\mathrm{There}'\mathrm{s}\:\mathrm{no}\:\mathrm{proof}\:\mathrm{as}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{matter}\:\mathrm{of} \\ $$$$\mathrm{definition}.\:\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{explain}\:\mathrm{but}\:\mathrm{maybe} \\ $$$$\mathrm{you}\:\mathrm{need}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{more}\:\mathrm{before}\:\mathrm{you}\:\mathrm{can}…
Question Number 192597 by York12 last updated on 22/May/23 $${a}_{\mathrm{1}\:} \:,\:{a}_{\mathrm{2}} \:,\:{a}_{\mathrm{3}\:} \:,\:….\:,\:{a}_{{n}} \:{is}\:{a}\:{sequence}\:{satifies}\:{that} \\ $$$${a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \:{for}\:{n}\:\geq\:\mathrm{1}.\:{suppose}\:{the}\:{sum}\: \\ $$$${of}\:{the}\:{first}\:\mathrm{999}\:{terms}\:=\:\mathrm{1003}\:{and}\:{the}\:{sum} \\ $$$${of}\:{the}\:{first}\:\mathrm{1003}\:{terms}\:=\:−\mathrm{999}\:{find}\:{the}\: \\ $$$${sum}\:{of}\:{the}\:{first}\:\mathrm{2002}\:{terms}.…
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Question Number 127056 by Khalmohmmad last updated on 26/Dec/20 Answered by Olaf last updated on 26/Dec/20 $$\frac{\mathrm{25}{e}}{\mathrm{0}^{−} }\:=\:−\infty \\ $$ Terms of Service Privacy Policy…
Question Number 192590 by York12 last updated on 22/May/23 $$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{{m}^{\mathrm{2}} {n}}{\mathrm{3}^{{m}} \left(\mathrm{3}^{{n}} .{m}+\mathrm{3}^{{m}} .{n}\right)}\right]=\lambda\: \\ $$$${find}\:\lambda. \\ $$$$ \\…