Question Number 205817 by hardmath last updated on 31/Mar/24 $$\mathrm{2}^{\:\boldsymbol{\mathrm{x}}\:+\:\boldsymbol{\mathrm{log}}_{\mathrm{2}} \:\mathrm{3}} \:=\:\mathrm{12}\:\:\Rightarrow\:\:\mathrm{find}:\:\:\mathrm{x}\:=\:? \\ $$ Answered by A5T last updated on 31/Mar/24 $$\left(\mathrm{2}^{{x}} \right)\left(\mathrm{2}^{{log}_{\mathrm{2}} \mathrm{3}} \right)=\mathrm{12}\Rightarrow\left(\mathrm{2}^{{x}}…
Question Number 205772 by mr W last updated on 30/Mar/24 Answered by MM42 last updated on 30/Mar/24 $${S}_{{n}} =\left(\sqrt{\mathrm{1}×\mathrm{2}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\mathrm{2}×\mathrm{3}}−\sqrt{\mathrm{1}×\mathrm{2}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\mathrm{3}×\mathrm{4}}−\sqrt{\mathrm{2}×\mathrm{3}}−\mathrm{1}\right) \\ $$$$\vdots…
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Question Number 205767 by lmcp1203 last updated on 30/Mar/24 $$ \\ $$$${a},{b},{c}\:\in\Re^{+} \:\: \\ $$$${a}+{b}+{c}=\mathrm{1} \\ $$$$\:\:\:{a}^{\mathrm{2}} /\left(\mathrm{1}+{b}+{c}\right)\:+\:{b}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{c}\right)\:\:+\:{c}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{b}\right)\geqslant{k} \\ $$$${find}\:\:\:{k}\:{max}. \\ $$$${hint}\::\:{inequality}\:{cauchy}\:{schwarz} \\…
Question Number 205770 by hardmath last updated on 30/Mar/24 $$\mathrm{If}\:\:\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{xyz}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\left(\sqrt{\mathrm{2}}\mathrm{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{y}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{yz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{yz}\right)}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 205746 by Satyam1234 last updated on 29/Mar/24 Answered by A5T last updated on 29/Mar/24 $${C}\neq{J}+\mathrm{2}\Rightarrow{C}=\mathrm{3}{F} \\ $$$${A}\geqslant\mathrm{1};{A}=\mathrm{1}\Rightarrow{B}=\mathrm{3}\Rightarrow{I}=\mathrm{7}\Rightarrow{H}=\mathrm{4}\Rightarrow{A}=\mathrm{8}\:{X} \\ $$$${A}=\mathrm{2}\Rightarrow{B}=\mathrm{6}\Rightarrow{E}=\mathrm{2}\:{X} \\ $$$${A}=\mathrm{3}\Rightarrow{B}=\mathrm{9}\:{X}\left(\mathrm{6}\Rightarrow{B}<\mathrm{5}\:{or}\:{A}\geqslant\mathrm{4}\right)\:{or}\:{G}=\mathrm{0} \\ $$$$\:\left({G}=\mathrm{0}\Rightarrow{I}=\mathrm{3}\Rightarrow{D}=\mathrm{6}\Rightarrow{E}=\mathrm{18}\right){X}…
Question Number 205726 by hardmath last updated on 28/Mar/24 $$ \\ $$101 is chosen arbitrarily from the numbers 1,2,3,…,199,200. Prove that two of these selected…
Question Number 205733 by hardmath last updated on 28/Mar/24 $$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}^{+} \:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{6} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4a}^{\mathrm{2}} −\mathrm{9a}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4b}^{\mathrm{2}} −\mathrm{9b}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4c}^{\mathrm{2}} −\mathrm{9c}\:+\:\mathrm{6}}\:\leqslant\:\mathrm{0} \\ $$ Answered by…