Question Number 182379 by Rasheed.Sindhi last updated on 08/Dec/22 $$\mathrm{Can}\:\mathrm{we}\:\mathrm{show}\:\mathrm{a}+\mathrm{b}<\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \\ $$$$\forall\:\mathrm{a},\mathrm{b}\in\mathbb{N} \\ $$ Commented by Frix last updated on 08/Dec/22 $$\mathrm{Check}\:\mathrm{it}\:\mathrm{for}\:{a}\in\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2}\right\} \\ $$…
Given-a-gt-b-gt-0-a-amp-b-real-number-such-that-a-2-ab-b-2-7-and-a-ab-b-1-find-the-value-of-a-2-b-2-
Question Number 116824 by bemath last updated on 07/Oct/20 $$\mathrm{Given}\:\mathrm{a}>\mathrm{b}>\mathrm{0}\:,\:\mathrm{a\&b}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} =\mathrm{7}\:\mathrm{and}\:\mathrm{a}−\mathrm{ab}+\mathrm{b}=−\mathrm{1}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \\ $$ Answered by 1549442205PVT last updated on…
Question Number 182336 by Shrinava last updated on 07/Dec/22 Answered by mr W last updated on 08/Dec/22 Commented by Acem last updated on 08/Dec/22 $${Your}\:{old}\:{and}\:{simpler}\:{method}\:{was}\:{also}\:{clear}…
Question Number 116780 by ajfour last updated on 06/Oct/20 Commented by ajfour last updated on 06/Oct/20 $${The}\:{curve}\:{in}\:{black}\:{is}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{x}}−\boldsymbol{{c}} \\ $$$$\:\:\:\left({c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)\:.\:{Find}\:{the}\:{radius}\:{of}\:{circle}. \\ $$ Answered by mr…
Question Number 51235 by peter frank last updated on 25/Dec/18 $${Without}\:{using}\:{tables}, \\ $$$${find}\:{tha}\:{value}\:{of} \\ $$$$\frac{\left(\sqrt{\mathrm{5}}\:+\mathrm{2}\right)^{\mathrm{6}} −\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)^{\mathrm{6}} }{\mathrm{8}\sqrt{\mathrm{5}}\:}\:\: \\ $$ Answered by mr W last updated…
Question Number 51227 by mr W last updated on 25/Dec/18 $${How}\:{many}\:{odd}\:{numbers}\:{with}\:{different} \\ $$$${digits}\:{are}\:{there}\:{from}\:\mathrm{2019}\:{to}\:\mathrm{9102}? \\ $$ Commented by mr W last updated on 25/Dec/18 $$\mathrm{1817}? \\…
Question Number 116759 by bemath last updated on 06/Oct/20 Answered by bobhans last updated on 06/Oct/20 $$=\underset{\mathrm{n}=\mathrm{0}} {\overset{\mathrm{21}} {\prod}}\left(\mathrm{2030}^{\mathrm{2}} −\left(\mathrm{2040}−\mathrm{n}\right)^{\mathrm{2}} \right) \\ $$$$=\underset{\mathrm{n}=\mathrm{0}} {\overset{\mathrm{21}} {\prod}}\left(\mathrm{2030}+\left(\mathrm{2040}−\mathrm{n}\right)\left(\mathrm{2030}−\left(\mathrm{2040}−\mathrm{n}\right)\right)\right.…
Question Number 182272 by SEKRET last updated on 06/Dec/22 $$\:\:\:\:\mathrm{1}\leqslant\boldsymbol{\mathrm{a}}\leqslant\mathrm{37} \\ $$$$\:\:\:\:\mathrm{1}\leqslant\boldsymbol{\mathrm{b}}\leqslant\mathrm{37} \\ $$$$\:\:\:\mathrm{1}+\mathrm{7a}+\mathrm{8}\boldsymbol{\mathrm{b}}\:+\mathrm{19}\boldsymbol{\mathrm{ab}}\:\:=\:\mathrm{0}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right)\: \\ $$$$\:\:\boldsymbol{\mathrm{a}}\:\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{b}}\:\:\boldsymbol{\mathrm{natural}}\:\:\boldsymbol{\mathrm{nambers}} \\ $$$$\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}_{\mathrm{1}} ;\boldsymbol{\mathrm{b}}_{\mathrm{1}} \right)\:\left(\boldsymbol{\mathrm{a}}_{\mathrm{2}} ;\boldsymbol{\mathrm{b}}_{\mathrm{2}} \right)……\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ;\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} \right) \\…
Question Number 116722 by ayenisamuel last updated on 06/Oct/20 $${find}\:{the}\:{range}\:{of}\:{values}\:{of}\:{k}\:{for} \\ $$$${which}\:{the}\:{equation}\:{e}^{{x}} −\mathrm{5}={k}\:{has}\:{no} \\ $$$${solution} \\ $$ Answered by Rio Michael last updated on 06/Oct/20…
Question Number 182247 by Shrinava last updated on 06/Dec/22 $$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{9}\:+\:\mathrm{4}\:\sqrt{\mathrm{5}}}\:−\:\sqrt{\mathrm{9}\:−\:\mathrm{4}\:\sqrt{\mathrm{5}}}\:=\:? \\ $$ Answered by mr W last updated on 06/Dec/22 $$\mathrm{9}\pm\mathrm{4}\sqrt{\mathrm{5}}=\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \pm\mathrm{4}\sqrt{\mathrm{5}}+\mathrm{2}^{\mathrm{2}} =\left(\sqrt{\mathrm{5}}\pm\mathrm{2}\right)^{\mathrm{2}}…