Question Number 99094 by bemath last updated on 18/Jun/20 $$\mathrm{find}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\mathrm{9}\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\mathrm{9}\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\mathrm{9}\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+…}}}}− \\ $$$$\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}+\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{}}}}…}}}}\: \\ $$ Answered by bobhans last updated on 18/Jun/20 $$\mathrm{y}\:=\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\mathrm{9y}}\:\Rightarrow\mathrm{y}^{\mathrm{3}} −\mathrm{9y}−\mathrm{9}\:=\:\mathrm{0} \\ $$$$\mathrm{x}=\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}+\mathrm{x}}}}\:\Rightarrow\:\mathrm{x}^{\mathrm{3}}…
Question Number 99089 by bramlex last updated on 18/Jun/20 $$\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+…}}}}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−…}}}} \\ $$ Answered by MJS last updated on 18/Jun/20 $${a}=\sqrt[{\mathrm{3}}]{\mathrm{9}+\sqrt[{\mathrm{3}}]{\mathrm{9}+…}} \\ $$$${a}^{\mathrm{3}} =\mathrm{9}+{a} \\ $$$${a}^{\mathrm{3}}…
Question Number 33518 by alekan251 last updated on 18/Apr/18 $$\alpha^{\mathrm{4}} +\beta^{\mathrm{4}\:} {solve}\:{please} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33515 by alekan251 last updated on 18/Apr/18 $${expand}\:\alpha^{\mathrm{4}} +\beta^{\beta\:\:} {please} \\ $$ Commented by math khazana by abdo last updated on 18/Apr/18 $${if}\:{you}\:{mean}\:\:\alpha^{\mathrm{4}}…
Question Number 99045 by bobhans last updated on 18/Jun/20 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{10}}\\{\mathrm{x}^{\mathrm{2}} −\mathrm{5xy}+\mathrm{6y}^{\mathrm{2}} \:=\:\mathrm{0}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{x}\:\&\mathrm{y}\: \\ $$ Answered by bemath last updated on 18/Jun/20…
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