Question Number 173054 by mathlove last updated on 06/Jul/22 Commented by Shrinava last updated on 06/Jul/22 $$\mathrm{a}+\mathrm{1}=\mathrm{x}\:\:,\:\:\mathrm{b}+\mathrm{1}=\mathrm{y}\:\:,\:\:\mathrm{c}+\mathrm{1}=\mathrm{z} \\ $$$$\mathrm{then}\:\:\:\mathrm{xyz}=\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=−\mathrm{6}\:\:\:\left(\mathrm{i}\right) \\ $$$$\Rightarrow\:\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{y}+\mathrm{2}\right)\left(\mathrm{z}+\mathrm{2}\right)=−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{4x}+\mathrm{4y}+\mathrm{4z}+\mathrm{2xy}+\mathrm{2yz}+\mathrm{2zx}=−\mathrm{12}…
Question Number 173048 by Shrinava last updated on 05/Jul/22 Answered by MJS_new last updated on 06/Jul/22 $$\mathrm{obviously}\:{x}={y}={z}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 173051 by mr W last updated on 05/Jul/22 $${solve} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{6}\right)=\mathrm{5}{x}^{\mathrm{2}} \\ $$ Answered by dragan91 last updated on 06/Jul/22 $$\left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{12}\right)\left({x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\right)−\mathrm{5}{x}^{\mathrm{2}}…
Question Number 173042 by mathlove last updated on 05/Jul/22 Answered by Rasheed.Sindhi last updated on 05/Jul/22 $$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{3} \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\…
Question Number 173028 by mnjuly1970 last updated on 05/Jul/22 Answered by MJS_new last updated on 05/Jul/22 $${x}=\mathrm{0} \\ $$ Answered by MJS_new last updated on…
Question Number 41957 by behi83417@gmail.com last updated on 15/Aug/18 Commented by MJS last updated on 16/Aug/18 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{any}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{all}\:\mathrm{variables}\:\in\mathbb{R} \\ $$$$\mathrm{in}\:\mathbb{C}\:\mathrm{we}\:\mathrm{can}\:\mathrm{freely}\:\mathrm{choose}\:\mathrm{3}\:\mathrm{variables}\:\mathrm{and}\:\mathrm{solve} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{remaining},\:\mathrm{so}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{unique}\:\mathrm{solution} \\ $$ Answered by…
Question Number 173027 by Shrinava last updated on 05/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 173024 by Shrinava last updated on 05/Jul/22 $$\mathrm{Find}\:\:\mathrm{A},\mathrm{B}\subset\mathbb{N}^{\ast} \\ $$$$\mathrm{Such}\:\mathrm{that}\:\:\mathrm{A}\cap\mathrm{B}=\varnothing\:,\:\mathrm{cardB}=\mathrm{cardA}+\mathrm{1}\geqslant\mathrm{8} \\ $$$$\mathrm{and}\:\:\underset{\boldsymbol{\mathrm{x}}\in\boldsymbol{\mathrm{A}}} {\sum}\:\mathrm{x}^{\mathrm{3}} \:=\:\underset{\boldsymbol{\mathrm{x}}\in\boldsymbol{\mathrm{B}}} {\sum}\:\mathrm{x}^{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 107486 by bemath last updated on 11/Aug/20 $$\:\:\:\:\:\:\nparallel\mathcal{B}{e}\mathcal{M}{ath}\nparallel \\ $$$$\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{5}}{\mathrm{25}}+\frac{\mathrm{8}}{\mathrm{125}}+\frac{\mathrm{11}}{\mathrm{625}}+\frac{\mathrm{14}}{\mathrm{3125}}+…\:=\:? \\ $$ Commented by bemath last updated on 11/Aug/20 $${thank}\:{you}\:{all}.\:{cooll}.. \\ $$ Answered…
Question Number 107483 by john santu last updated on 11/Aug/20 $$\:\:\:\:\:\:\:\:\divideontimes\mathcal{JS}\divideontimes \\ $$$$\:\:\:\:\:\mid\mathrm{1}+\frac{\mathrm{1}}{{x}}\:\mid−\mid{x}−\mathrm{3}\mid\:>\:\mathrm{2}\: \\ $$$${find}\:{solution}\:{set}. \\ $$$$\left({A}\right)\:\mathrm{3}−\sqrt{\mathrm{10}}\:<\:{x}\:<\:\mathrm{2}−\sqrt{\mathrm{3}}\:;\:{x}\neq\mathrm{0} \\ $$$$\left({B}\right)\:\mathrm{3}−\sqrt{\mathrm{10}}\:<\:{x}\:<\:\mathrm{3}+\sqrt{\mathrm{10}}\:;\:{x}\neq\mathrm{0} \\ $$$$\left({C}\right)\:\mathrm{3}−\sqrt{\mathrm{10}}\:<\:{x}\:<\:\mathrm{2}+\sqrt{\mathrm{10}}\:;\:{x}\neq\mathrm{0} \\ $$$$\left({D}\right)\:\mathrm{2}+\sqrt{\mathrm{10}}\:<\:{x}\:<\:\mathrm{3}+\sqrt{\mathrm{10}}\:;\:{x}\neq\mathrm{0} \\ $$$$\left({E}\right)\:{none}\:{of}\:{these}\:…