Question Number 164568 by leonhard77 last updated on 19/Jan/22 $$\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−{x}}}\:>\:\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$ Answered by TheSupreme last updated on 19/Jan/22 $${x}<\mathrm{2},\:{x}\neq\mathrm{1} \\ $$$$\begin{cases}{\frac{\mathrm{1}}{{x}−\mathrm{1}}<\mathrm{0}\:\cup\:\begin{cases}{\frac{\mathrm{1}}{{x}−\mathrm{1}}>\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{2}−{x}}>\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }}\end{cases}}\\{}\end{cases} \\ $$$$\begin{cases}{{x}<\mathrm{1}\:\cup\:\begin{cases}{{x}>\mathrm{1}}\\{{x}^{\mathrm{2}}…
Question Number 164576 by mathlove last updated on 19/Jan/22 Answered by Rasheed.Sindhi last updated on 19/Jan/22 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:={a}\Rightarrow\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}\centerdot\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}} \\ $$$$\frac{\mathrm{1}}{{a}}=\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\: \\ $$$$\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{{x}} +\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} \:=\mathrm{4} \\ $$$$\Rightarrow{a}^{{x}}…
Question Number 33496 by malwaan last updated on 17/Apr/18 $$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{e}^{\mathrm{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$ Answered by MJS last updated on 17/Apr/18 $$\mathrm{just}\:\mathrm{use}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:{z}\in\mathbb{C}: \\ $$$${z}={r}\mathrm{e}^{\varphi\mathrm{i}}…
Question Number 33473 by 33 last updated on 17/Apr/18 $$\:{e}^{{i}\pi\:} =\:−\mathrm{1} \\ $$$${squaring}\:{both}\:{sides} \\ $$$${e}^{\mathrm{2}\pi{i}} \:=\:\mathrm{1}\:=\:{e}^{\mathrm{0}} \\ $$$${comparing}\:{powers} \\ $$$$\mathrm{2}\pi{i}\:=\:\mathrm{0} \\ $$$$\:\pi\:=\:\mathrm{0}\:{or}\:{i}\:=\:\mathrm{0}\:??? \\ $$ Commented…
Question Number 164542 by cortano1 last updated on 18/Jan/22 $$\:\:\:\:\:\:\:\frac{{x}^{\mathrm{2}} −{x}}{\mathrm{3}−\mathrm{2}{x}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:<\:\frac{\mathrm{5}}{−\mathrm{3}{x}+{x}^{\mathrm{2}} −\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 164543 by cortano1 last updated on 18/Jan/22 $$\:\:\frac{\mathrm{4}\sqrt{\mathrm{1}−{x}}}{{x}}\:+\frac{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{1}−{x}}\:=\:\mathrm{3}\sqrt{\mathrm{3}} \\ $$ Answered by leonhard77 last updated on 19/Jan/22 $$\:\left(\mathrm{1}\right)\:\mathrm{1}−{x}>\mathrm{0}\:\cap\:\mathrm{2}{x}−\mathrm{1}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}<\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{let}\:\sqrt{{x}}\:=\mathrm{cos}\:\:{t} \\…
Question Number 33468 by NECx last updated on 17/Apr/18 $${The}\:{set}\:{of}\:{integers}\:{that}\:{satisfies} \\ $$$$\mathrm{5}>\mid{n}−\mathrm{2}\mid\geqslant\mid{n}+\mathrm{1}\mid\:{is} \\ $$ Answered by MJS last updated on 17/Apr/18 $$\mathrm{25}>\left({n}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} −\mathrm{4}{n}−\mathrm{21}<\mathrm{0}…
Question Number 99005 by bobhans last updated on 18/Jun/20 $$\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{mn}\left(\mathrm{m}+\mathrm{n}\right)}\:?\: \\ $$ Answered by maths mind last updated on 18/Jun/20 $$\underset{{m}\geqslant\mathrm{1}}…
Question Number 164533 by alephzero last updated on 18/Jan/22 $$\mathrm{Prove},\:\mathrm{that} \\ $$$$\left.\mathrm{1}\right)\:\underset{{k}=\mathrm{1}} {\overset{\alpha} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}\:=\:\frac{\alpha}{\alpha+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+{x}\right)}\:=\:\mathrm{0} \\ $$ Answered by amin96 last…
Question Number 98983 by M±th+et+s last updated on 17/Jun/20 $${let}\:{a},{b},{c}\:{be}\:{positive}\:{real}\:{numbers}\:{such} \\ $$$${that}\:{ab}+{bc}+{ac}=\mathrm{3}\: \\ $$$${prove}\:{the}\:{inquality} \\ $$$$ \\ $$$$\frac{{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} +{bc}}+\frac{{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} +{ac}}+\frac{{c}\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}}…