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Question Number 164970 by Zaynal last updated on 24/Jan/22 $$\:\:\lfloor\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{1}} }{\frac{\boldsymbol{{x}}}{\mathrm{10}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{2}} }{\frac{\boldsymbol{{x}}}{\mathrm{9}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{3}} }{\frac{\boldsymbol{{x}}}{\mathrm{8}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\boldsymbol{{x}}\:\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{4}} }{\frac{\boldsymbol{{x}}}{\mathrm{7}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{5}} }{\frac{\boldsymbol{{x}}}{\mathrm{6}−\mathrm{3}^{\boldsymbol{{x}}} }}\:\geqslant\:\frac{\mathrm{1}}{\frac{\mathrm{25}}{\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}\:\left(\frac{\mathrm{1}}{\mathrm{25}}\right)} } }}\rfloor \\ $$$$\:\:\:\left\{\mathrm{Z}.\mathrm{A}\right\} \\…
Question Number 99430 by I want to learn more last updated on 20/Jun/20 Answered by mr W last updated on 20/Jun/20 $${f}\left({x}\right)={x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{2}} −\frac{\mathrm{44}}{\mathrm{9}}{x}−\frac{\mathrm{40}}{\mathrm{9}}=\mathrm{0} \\…
Question Number 164966 by Zaynal last updated on 24/Jan/22 $$\:\:\:\:\:\lfloor\left(\frac{\mathrm{5}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{3}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{2}}{\boldsymbol{{x}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)\:\bullet\:\left(\frac{\boldsymbol{{x}}}{\mathrm{1}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{2}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{3}}\:\:−\:\frac{\boldsymbol{{x}}}{\mathrm{4}}\:−\:\frac{\boldsymbol{{x}}}{\mathrm{5}}\:\right)^{\mathrm{2}} >\:\frac{\mathrm{1}}{\mathrm{15}}\rfloor \\ $$$$\:\:\:\left\{\mathrm{za}\right\} \\ $$ Answered by alephzero last updated on 24/Jan/22 $$\frac{\mathrm{15}}{{x}}\:×\:\frac{\mathrm{17}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{60}^{\mathrm{2}}…
Question Number 33880 by pieroo last updated on 26/Apr/18 Commented by pieroo last updated on 27/Apr/18 $$\mathrm{Please},\:\mathrm{i}\:\mathrm{really}\:\mathrm{need}\:\mathrm{help}\:\mathrm{with}\:\mathrm{the}\:\mathrm{above} \\ $$ Answered by math1967 last updated on…
Question Number 33876 by CaesarMedellin last updated on 26/Apr/18 $${how}\:{can}\:{i}\:{write}\:{out}\:\mathrm{6}\:{to}\:{thethird}\:{power}\:{divided}\:{by}\:\mathrm{2}\:{to}\:{the}\:{fourth}\:{power}\:{pleasr}\:{help}\:{me}\:{write}\:{out}\:{this}\:{equation}\: \\ $$ Commented by prakash jain last updated on 26/Apr/18 $$\frac{\mathrm{6}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{4}} } \\ $$…
Question Number 164945 by mnjuly1970 last updated on 23/Jan/22 $$ \\ $$$$\:\:\mathrm{I}{f}\:\:\:\begin{cases}{\:\:{sin}\:\left(\:\mathrm{3}\theta\:\right)\:+\:{cos}\:\left(\:\mathrm{3}\theta\:\right)\:=\:{x}}\\{\:\:\:\:\:{sin}\left(\:\theta\:\right)\:+\:{cos}\:\left(\theta\:\right)\:=\:{y}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:,\:{find}\:\:{a}\:{relationship}\: \\ $$$$\:\:\:\:\:\:\:\:{between}\:\:\:{x}\:\:,\:\:{y}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{indepentent}\:{of}\:,\:\:\:\theta\:. \\ $$ Answered by mr W last…
Question Number 164924 by cortano1 last updated on 23/Jan/22 Commented by bobhans last updated on 23/Jan/22 $$\:\left(\mathrm{1}\right)\:\mathrm{m}^{\mathrm{3}} =\mathrm{2020}+\sqrt{\mathrm{2019}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{n}^{\mathrm{3}} \:=\:\mathrm{2017}\:+\sqrt{\mathrm{2019}} \\ $$$$\:\:\:\:\:\:\:\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} \:=\:\mathrm{3}\:;\:\mathrm{3}+\mathrm{n}^{\mathrm{3}}…
Question Number 164926 by Zaynal last updated on 23/Jan/22 $$\:\:\:\:\:\:\:\:\:\:\:\lfloor\boldsymbol{\mathrm{Given}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{numbers}}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}},\boldsymbol{{d}}\rceil \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}; \\ $$$$\:\:\:\:\:\:\:\:\lfloor\frac{\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{b}}^{\mathrm{3}} +\boldsymbol{{c}}^{\mathrm{3}} }{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}}\:+\:\frac{\boldsymbol{{b}}^{\mathrm{3}} +\boldsymbol{{c}}^{\mathrm{3}} +\boldsymbol{{d}}^{\mathrm{3}} }{\boldsymbol{{b}}+\boldsymbol{{c}}+\boldsymbol{{d}}}\:+\:\frac{\boldsymbol{{c}}^{\mathrm{3}} +\boldsymbol{{d}}^{\mathrm{3}} +\boldsymbol{{a}}^{\mathrm{3}} }{\boldsymbol{{c}}+\boldsymbol{{d}}+\boldsymbol{{a}}}\:+\:\frac{\boldsymbol{{d}}^{\mathrm{3}} +\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{b}}^{\mathrm{3}}…