Question Number 188535 by Rupesh123 last updated on 03/Mar/23 Answered by mr W last updated on 03/Mar/23 $${due}\:{to}\:{symmetry} \\ $$$${at}\:{extremum} \\ $$$${a}={b}={c}={x}>\mathrm{0} \\ $$$${S}=\frac{\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{3}} }{{x}}=\frac{\mathrm{3}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{{x}}…
Question Number 188515 by mnjuly1970 last updated on 02/Mar/23 $$ \\ $$$$\:\:\:\:\:{in}\:\:{A}\overset{\Delta} {{B}C}\:\::\:\:\:{a}=\mathrm{3}\:\:,\:\:{b}=\mathrm{6}\:\:,\:\:{c}=\mathrm{7} \\ $$$$\:\:\: \\ $$$$\: \\ $$$$\:\:\:\:{find}\:\:{the}\:{value}\:\:{of}\:: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:{E}\:=\:\left({a}+{b}\:\right){cos}\left({C}\right)\:+\:\left({b}+{c}\right){cos}\left({A}\right)+\:\left({a}+{c}\:\right){cos}\left({B}\right)=?\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:…
Question Number 188508 by Rupesh123 last updated on 02/Mar/23 Answered by som(math1967) last updated on 02/Mar/23 $$\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)=\mathrm{0}…
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Question Number 57434 by Tawa1 last updated on 04/Apr/19 $$\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{product}\:\mathrm{operator} \\ $$$$\:\:\mathrm{e}.\mathrm{g}\:\:\:\:\:\mathrm{product}\:\mathrm{of}\:\:\:\:\:\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}\:…\:\:\left[\mathrm{1},\:\mathrm{infinity}\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 122957 by I want to learn more last updated on 21/Nov/20 Commented by mr W last updated on 21/Nov/20 $${see}\:{Q}\mathrm{74970} \\ $$ Answered…
Question Number 122950 by bemath last updated on 21/Nov/20 $$\:\:\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}+…\sqrt{\mathrm{1}+\mathrm{2014}+\sqrt{\mathrm{1}+\mathrm{2015}.\mathrm{2016}}}}}}\:=\:? \\ $$ Commented by Dwaipayan Shikari last updated on 21/Nov/20 $$\mathrm{2} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}+\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{…}}}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{1}+\mathrm{3}}=\mathrm{2}…
Question Number 188482 by mathlove last updated on 02/Mar/23 $$\mathrm{512}{x}^{\mathrm{1}−{x}^{−\mathrm{3}} } =−\mathrm{1} \\ $$$${find}\:\:{volue}\:\:{of}\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({x}^{\mathrm{2}} \right)^{{n}} =? \\ $$ Answered by mr W last…
Question Number 122897 by Study last updated on 20/Nov/20 Answered by Dwaipayan Shikari last updated on 20/Nov/20 $$\mathrm{9}+\mathrm{9}+\mathrm{15}×\mathrm{3}=\mathrm{63} \\ $$$${or} \\ $$$$\mathrm{9}+\mathrm{9}+\mathrm{15}^{\mathrm{3}} =\mathrm{3383} \\ $$$${Many}\:{more}\:{possibilities}..…
Question Number 188430 by mnjuly1970 last updated on 01/Mar/23 $$ \\ $$$$\:\:\:\:\mathrm{2}\lfloor\:{x}\:\rfloor\:−\:\lfloor\:−{x}\:\rfloor\:=\mathrm{4} \\ $$$$\:\:\:−−−− \\ $$$$\:\:{if}\:\:{x}\in\mathbb{Z}\:\Rightarrow\:\:\mathrm{2}{x}\:+{x}\:=\:\mathrm{4}\:\Rightarrow\:{x}=\frac{\mathrm{4}}{\mathrm{3}}\:\:,{impossible} \\ $$$$\:\:{if}\:{x}\notin\:\mathbb{Z}\:\overset{\lfloor−{x}\rfloor=−\lfloor{x}\rfloor−\mathrm{1}} {\Rightarrow}\mathrm{2}\lfloor{x}\rfloor+\lfloor{x}\rfloor=\mathrm{3} \\ $$$$\:\:\:\:\:\Rightarrow\:\lfloor\:{x}\:\rfloor=\:\mathrm{1}\:\Rightarrow\:\:\mathrm{1}\leqslant\:{x}\:<\:\mathrm{2}\:\:\:\:\overset{{x}\neq\mathrm{1}} {\Rightarrow}\:{x}\in\:\left(\mathrm{1}\:,\:\mathrm{2}\right)\:\:\:\checkmark \\ $$$$ \\…