Question Number 55685 by tm888 last updated on 02/Mar/19 $${proof}\:{that}\: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}}=\mathrm{2015}\:{has}\:{exactly}\:{n}\:{real}\: \\ $$$${roots}.{o}<{a}_{\mathrm{1}} ….<{a}_{{n}} \\ $$ Answered by mr W…
Question Number 55673 by gunawan last updated on 01/Mar/19 $${f}\left({z}\right)={z}\:\mathrm{Re}\left({z}\right)+\bar {{z}}\:\mathrm{Im}\left({z}\right)\:+\bar {{z}}\: \\ $$$${f}'\left({z}_{\mathrm{0}} \right)=… \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 55671 by gunawan last updated on 01/Mar/19 $$\mathrm{Let}\:\mathrm{z}\:\in\:\mathbb{C}\:,\:\mathrm{so}\:\mid\mathrm{1}+{z}^{\mathrm{2}} \mid<\mathrm{1}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{2}\mid\mathrm{1}+{z}^{\mathrm{2}} \mid\geqslant\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 186732 by mr W last updated on 09/Feb/23 Answered by mr W last updated on 09/Feb/23 $${let}'{s}\:{generally}\:{find}\:{the}\:{maximum}\:{of} \\ $$$${S}={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 121170 by pooooop last updated on 05/Nov/20 $$\:\:\:\:\:\:\left(\boldsymbol{{a}};\boldsymbol{{b}};\boldsymbol{{c}}\right)\in\forall \\ $$$$\:\boldsymbol{{a}}^{\boldsymbol{{a}}} \centerdot\:\boldsymbol{{b}}^{\boldsymbol{{b}}} \centerdot\:\boldsymbol{{c}}^{\boldsymbol{{c}}} \geqslant\left(\boldsymbol{{abc}}\right)^{\frac{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}}{\mathrm{3}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186701 by Mingma last updated on 08/Feb/23 Commented by aba last updated on 08/Feb/23 $$\frac{\mathrm{4}}{\mathrm{5}} \\ $$ Commented by Mingma last updated on…
Question Number 55625 by gunawan last updated on 28/Feb/19 Commented by kaivan.ahmadi last updated on 28/Feb/19 $${the}\:{image}\:{is}\:{not}\:{clear} \\ $$ Commented by gunawan last updated on…
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Question Number 186690 by mathlove last updated on 08/Feb/23 Answered by a.lgnaoui last updated on 08/Feb/23 $$\mathrm{x}×\mathrm{2}^{\mathrm{x}^{\mathrm{2}} } =\mathrm{4}\:\:\:\:\left(\mathrm{x}=\mathrm{a}^{\mathrm{2}} \right) \\ $$$$\mathrm{x}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{x}^{\mathrm{2}} } =\mathrm{4x}…