Question Number 99171 by Quvonchbek last updated on 19/Jun/20 Commented by MJS last updated on 19/Jun/20 $${A}={B}\:\Rightarrow\:\frac{{A}}{{B}}=\mathrm{1}\:\Rightarrow\:\left(\frac{{A}}{{B}}\right)^{\mathrm{2020}} =\mathrm{1} \\ $$ Commented by Quvonchbek last updated…
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Question Number 99117 by M±th+et+s last updated on 18/Jun/20 $${let}\:{a},{b},{c}\:\in\mathbb{R}\:{determine}\:{the}\:{minimum} \\ $$$${value} \\ $$$$ \\ $$$$\frac{\mathrm{3}{a}}{{b}+{c}}+\frac{\mathrm{4}{b}}{{a}+{c}}+\frac{\mathrm{5}{c}}{{a}+{b}} \\ $$ Answered by MJS last updated on 19/Jun/20…
Question Number 164650 by cortano1 last updated on 20/Jan/22 $$\:\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{9}}\:−\sqrt[{\mathrm{3}}]{{x}−\mathrm{9}}\:=\:\mathrm{3}\: \\ $$$$\:{x}=? \\ $$ Answered by mr W last updated on 20/Jan/22 $$\left(\sqrt[{\mathrm{3}}]{\mathrm{9}+{x}}+\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}}\right)^{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} \\…
Question Number 33569 by Joel578 last updated on 19/Apr/18 $$\mathrm{Given}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c} \\ $$$$\mathrm{with}\:{a},\:{b},\:{c}\:\in\:\mathbb{R},\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:{x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:{x}_{\mathrm{3}} \:\in\:\mathbb{R} \\ $$$$\mathrm{Let}\:\lambda\:\mathrm{is}\:\mathrm{an}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{satisfied} \\ $$$${x}_{\mathrm{2}} \:−\:{x}_{\mathrm{1}} \:=\:\lambda \\ $$$${x}_{\mathrm{3}}…
Question Number 99094 by bemath last updated on 18/Jun/20 $$\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…