Question Number 195765 by SajaRashki last updated on 10/Aug/23 $${hello} \\ $$$$ \\ $$$$\:\begin{cases}{{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{18}}\\{{x}>\mathrm{1}}\end{cases}\:\:\Rightarrow\:\:\:{x}^{\mathrm{5}} −\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\:=\:? \\ $$ Answered by mr W last…
Question Number 195799 by sonukgindia last updated on 10/Aug/23 Answered by som(math1967) last updated on 11/Aug/23 $$\:\frac{\mathrm{1}}{{a}+\frac{\mathrm{1}}{{a}}}\:+\frac{\mathrm{1}}{{a}−\frac{\mathrm{1}}{{a}}}=\mathrm{2}{a} \\ $$$$\Rightarrow\:\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2}{a} \\ $$$$\Rightarrow{a}\left[\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\right]=\mathrm{2}{a}…
Question Number 195790 by justenspi last updated on 10/Aug/23 $${a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers}\:{and}\:{abc}\:=\mathrm{1} \\ $$$${prove}\:{that} \\ $$$$\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right)\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)\leqslant\mathrm{1} \\ $$ Commented by justenspi last updated on 10/Aug/23 $${Case}\left({I}\right)\left({a}−\mathrm{1}+\frac{\mathrm{1}}{{b}}\right),\left({b}−\mathrm{1}+\frac{\mathrm{1}}{{c}}\right),\left({c}−\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)>\mathrm{0} \\…
Question Number 195740 by universe last updated on 09/Aug/23 Answered by Frix last updated on 09/Aug/23 $$\mathrm{Triangle}\:\Rightarrow\:{a}+{b}>{c}\wedge{a}+{c}>{b}\wedge{b}+{c}>{a} \\ $$$$\mathrm{Let}\:{b}=\left({u}−{v}\right){a}\wedge{c}=\left({u}+{v}\right){a} \\ $$$$\Rightarrow\:{u}>\frac{\mathrm{1}}{\mathrm{2}}\wedge−\frac{\mathrm{1}}{\mathrm{2}}<{v}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left({u},\:{v}\right)=\frac{\mathrm{1}}{\mathrm{2}{u}}+\frac{{u}−{v}}{{u}+{v}+\mathrm{1}}+\frac{{u}+{v}}{{u}−{v}+\mathrm{1}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant{f}\left({u},\:{v}\right)<\mathrm{2}…
Question Number 195742 by sonukgindia last updated on 09/Aug/23 Answered by MM42 last updated on 09/Aug/23 $$\angle\alpha_{{i}} =\mathrm{120}\Rightarrow\angle{BAC}=\angle{BDF}=\mathrm{90} \\ $$$${AB}={AC}\Rightarrow\angle{C}=\angle{B}_{\mathrm{1},\mathrm{2}} =\mathrm{45}\Rightarrow\angle{B}_{\mathrm{2}} =\angle{B}_{\mathrm{3}} =\mathrm{15} \\ $$$$\frac{{x}}{{sin}\mathrm{15}}=\frac{{BE}}{{sin}\mathrm{90}}\:\:\:\&\:\:\:\frac{{y}}{{sin}\mathrm{15}}=\frac{{BE}}{{sin}\mathrm{60}}…
Question Number 195733 by York12 last updated on 09/Aug/23 $${prove}\:{that} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left[\frac{{B}_{\overset{\_} {{n}}} }{\left({n}−\mathrm{2}\right)!}\right]=\frac{{e}\left(\mathrm{3}−{e}\right)}{\left({e}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${where}\:{B}_{\overset{\_} {{n}}} \:{is}\:{the}\:{n}−\:{th}\:{bernouli}'{s}\:{number} \\ $$ Answered by…
Question Number 195732 by bbbbbbbb last updated on 09/Aug/23 $$\frac{\mathrm{500}!+\mathrm{499}!}{\mathrm{0}.\mathrm{002}}=? \\ $$$$\boldsymbol{\mathrm{plz}}\:\boldsymbol{\mathrm{soon}}\: \\ $$ Answered by HeferH last updated on 09/Aug/23 $$\frac{\mathrm{500}\centerdot\mathrm{499}!\:+\:\mathrm{499}!}{\frac{\mathrm{2}}{\mathrm{1000}}}\:=\:\frac{\mathrm{499}!\centerdot\mathrm{501}\centerdot\mathrm{1000}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{499}!\:\centerdot\mathrm{501}\centerdot\mathrm{500}}{\mathrm{1}}\:=\:\mathrm{501}!\: \\…
Question Number 195753 by Erico last updated on 09/Aug/23 $$\mathrm{Calculer}\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} {t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$ Answered by witcher3 last updated on 09/Aug/23 $$\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 195755 by sonukgindia last updated on 09/Aug/23 Answered by HeferH last updated on 09/Aug/23 $${x}=\frac{\mathrm{180}°}{\mathrm{7}} \\ $$ Commented by AROUNAMoussa last updated on…
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