Question Number 202298 by hardmath last updated on 24/Dec/23 $$\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{6x}\:+\:\mathrm{11}}\right)\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 24/Dec/23 $$\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{6x}\:+\:\mathrm{11}}\right)=\boldsymbol{\mathrm{max}}\left(\frac{\mathrm{5}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}}\right)…
Question Number 202290 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:=\:\frac{{a}^{\mathrm{2}\:} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{ax}\:+\:{by}\:+\:{cz}}\:. \\ $$ Answered by Rasheed.Sindhi last updated on 24/Dec/23 $$\mathrm{If}\:\frac{\mathrm{3}{a}\:−\:{b}}{{x}\:+\:{y}}\:=\:\frac{\mathrm{3}{b}\:−\:{c}}{{y}\:+\:{z}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{{z}\:+\:{x}}\:\mathrm{then}\:\mathrm{show}…
Question Number 202348 by York12 last updated on 24/Dec/23 $$ \\ $$$$\mathrm{Let}\:{a},{b},{c}\:\:\in\mathbb{R}^{+} \:,\:{a}+{b}+{c}=\mathrm{3}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality} \\ $$$$\frac{\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} }{{b}}+\frac{\left(\mathrm{2}{b}−\mathrm{3}\right)^{\mathrm{2}} }{{c}}+\frac{\left(\mathrm{2}{c}−\mathrm{3}\right)^{\mathrm{2}} }{{a}}\geqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}+\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{{b}+{c}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{c}+{a}} \\…
Question Number 202340 by hardmath last updated on 24/Dec/23 Answered by MATHEMATICSAM last updated on 25/Dec/23 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{2}}}\:=\:\sqrt{\mathrm{3}}\:−\:\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}\:+\:\sqrt{\mathrm{3}}}\:=\:\sqrt{\mathrm{4}}\:−\:\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}\:+\:\sqrt{\mathrm{4}}}\:=\:\sqrt{\mathrm{5}}\:−\:\sqrt{\mathrm{4}} \\ $$$$. \\ $$$$.…
Question Number 202324 by hardmath last updated on 24/Dec/23 $$\mathrm{Find}:\:\:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:…\:+\:\mathrm{16}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:…\:+\:\mathrm{8}} \\ $$ Answered by MATHEMATICSAM last updated on 24/Dec/23 $$\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:….\:+\:\mathrm{16}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:….\:+\:\mathrm{8}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{2}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:\frac{\mathrm{4}}{\mathrm{2}}\:+\:….\:+\:\frac{\mathrm{32}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{4}}{\mathrm{4}}\:+\:….\:+\:\frac{\mathrm{32}}{\mathrm{4}}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:….\:+\:\mathrm{32}\right]}{\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:….\:+\:\mathrm{32}\right]} \\…
Question Number 202325 by Ikbal last updated on 24/Dec/23 Answered by AST last updated on 24/Dec/23 $${Let}\:{x}={re}^{{i}\theta} ={r}\left({cos}\theta+{isin}\left(\theta\right)\right); \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{{e}^{−{i}\theta} }{{r}}=\frac{\mathrm{1}}{{r}}\left({cos}\left(\theta\right)−{isin}\left(\theta\right)\right) \\ $$$$\Rightarrow{x}^{{n}} ={r}^{{n}} \left[{cos}\left({n}\theta\right)+{isin}\left({n}\theta\right)\right]…
Question Number 202315 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:{x}\::\:{y}\::\:{z}\:=\:{a}\::\:{b}\::\:{c}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\right)^{\mathrm{3}} \:=\:\frac{{abc}}{{xyz}}\:. \\ $$ Answered by AST last updated on 24/Dec/23 $${x}={ka};{y}={kb};{z}={kc} \\ $$$$\left(\frac{{a}+{b}+{c}}{{x}+{y}+{z}}\right)^{\mathrm{3}}…
Question Number 202306 by hardmath last updated on 24/Dec/23 $$ \\ $$Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,… obtained by subtracting integer squares from…
Question Number 202307 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{Show}\:\mathrm{that}\:\frac{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:>\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }{{a}\:+\:{b}} \\ $$ Answered by aleks041103 last updated on 24/Dec/23…
Question Number 202258 by MATHEMATICSAM last updated on 23/Dec/23 $$\mathrm{If}\:\frac{{x}}{{a}}\:=\:\frac{{y}}{{b}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\frac{{x}^{\mathrm{3}} \:+\:\mathrm{3}{xy}^{\mathrm{2}} }{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} }\:=\:\frac{\:{y}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} {y}}{{b}^{\mathrm{3}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}}\:. \\ $$ Answered by AST…