Question Number 42671 by Tawa1 last updated on 31/Aug/18 $$\mathrm{If}\:\mathrm{pqr}\:=\:\mathrm{1} \\ $$$$\mathrm{Hence}\:\mathrm{evaluate}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{e}\:+\:\mathrm{f}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{f}\:+\:\mathrm{g}^{−\mathrm{1}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{g}\:+\:\mathrm{e}^{−\mathrm{1}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 173732 by mr W last updated on 17/Jul/22 $${solve}\:{for}\:{x}\in{R} \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}={x}^{\mathrm{2}} \\ $$ Commented by dragan91 last updated on 18/Jul/22 $$\mathrm{let}\:\mathrm{x}^{\mathrm{2}}…
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Question Number 173709 by dragan91 last updated on 16/Jul/22 $$\mathrm{In}\:\mathrm{triangle}\:\bigtriangleup\mathrm{ABC}\:\mathrm{prove}: \\ $$$$\mathrm{cos}\left(\measuredangle\frac{\mathrm{A}}{\mathrm{2}}\right)>\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\mathrm{2a}+\mathrm{b}+\mathrm{c}}{\:\sqrt{\mathrm{R}\left(\mathrm{R}+\mathrm{2r}_{\mathrm{a}} \right)}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 173706 by infinityaction last updated on 16/Jul/22 Commented by Frix last updated on 16/Jul/22 $$\mathrm{exact}\:\mathrm{solution}\:\mathrm{possible}\:\mathrm{but}\:\mathrm{not}\:\mathrm{useable} \\ $$$$\mathrm{shortest}\:\mathrm{way}\:\mathrm{I}\:\mathrm{found}\:\mathrm{put}\:{y}={cx}\:\Rightarrow\:\begin{cases}{{x}^{\mathrm{3}} ={T}_{\mathrm{1}} }\\{{x}^{\mathrm{3}} ={T}_{\mathrm{2}} }\end{cases} \\ $$$${T}_{\mathrm{1}}…
Question Number 173707 by dragan91 last updated on 16/Jul/22 $$\mathrm{Let}\:\mathrm{x},\mathrm{y}\in\mathbb{R}\:\mathrm{such}\:\mathrm{that}\:\mathrm{2x}^{\mathrm{2}} +\mathrm{3y}^{\mathrm{2}} =\mathrm{5} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{and} \\ $$$$\mathrm{maximum}\:\mathrm{of}\:\mathrm{expression}: \\ $$$$\mathrm{P}=\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} +\mathrm{x}−\mathrm{2y} \\ $$ Answered by a.lgnaoui…
Question Number 173687 by mnjuly1970 last updated on 16/Jul/22 $$ \\ $$$$\:\:\:{Q}\::\:\:\:\:\:\frac{{ln}\left({a}\:\right)}{{c}−{b}}\:=\frac{{ln}\left({b}\right)}{{a}−{c}}\:=\:\frac{{ln}\left({c}\right)}{{b}−{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:{a}^{\:{a}} .\:{b}^{\:{b}} .\:{c}^{\:{c}} \:=? \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last…
Question Number 173629 by mnjuly1970 last updated on 15/Jul/22 Commented by mnjuly1970 last updated on 15/Jul/22 $$\:\:\:\:{solve}\:\:{for}\:\:\:\mathbb{R}\:\:\:{a},{b},{c}\:?\:\Uparrow\Uparrow\Uparrow \\ $$ Commented by Tawa11 last updated on…
Question Number 42559 by Tawa1 last updated on 27/Aug/18 Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18 $$\left.\mathrm{15}\right){a}+\frac{\mathrm{1}}{{a}}=\left(\sqrt{{a}}\:−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} +\mathrm{2}\: \\ $$$${so}\:{a}+\frac{\mathrm{1}}{{a}}=\mathrm{2}\:{if}\:\left(\sqrt{{a}}\:−\frac{\mathrm{1}}{\:\sqrt{{a}}\:}\right)=\mathrm{0} \\ $$$${a}+\frac{\mathrm{1}}{{a}}>\mathrm{2}\:{if}\left(\sqrt{{a}}\:−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)>\mathrm{0} \\ $$$$ \\…