Question Number 157453 by MathSh last updated on 23/Oct/21 Answered by MJS_new last updated on 23/Oct/21 $$\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{6}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{2}\:\vee\:{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\vee\:{x}=\pm\mathrm{i}\:\vee\:{x}=−\frac{\mathrm{1}\pm\sqrt{\mathrm{23}}\mathrm{i}}{\mathrm{2}} \\ $$ Commented…
Question Number 157454 by tounghoungko last updated on 23/Oct/21 $$\:{Given}\:\mathrm{1}+\frac{\mathrm{3}}{{y}}+\frac{\mathrm{5}}{{y}^{\mathrm{2}} }+\frac{\mathrm{7}}{{y}^{\mathrm{3}} }+\frac{\mathrm{9}}{{y}^{\mathrm{4}} }+…=\:\mathrm{71} \\ $$$$\:{then}\:\mathrm{1}+\frac{\mathrm{4}}{{y}}+\frac{\mathrm{9}}{{y}^{\mathrm{2}} }+\frac{\mathrm{16}}{{y}^{\mathrm{3}} }+\frac{\mathrm{25}}{{y}^{\mathrm{4}} }+…\:=? \\ $$ Answered by FongXD last updated…
Question Number 157441 by bobhans last updated on 23/Oct/21 $$\:\left(\mathrm{7}\:\sqrt[{\mathrm{5}}]{\mathrm{6x}−\mathrm{10}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{49}}{\:\sqrt[{\mathrm{5}}]{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}\: \\ $$ Answered by MJS_new last updated on 23/Oct/21 $$\mathrm{49}\left(\mathrm{6}{x}−\mathrm{10}\right)^{\mathrm{2}/\mathrm{5}} =\mathrm{49}{x}^{\mathrm{2}/\mathrm{5}} \\ $$$$\left(\mathrm{6}{x}−\mathrm{10}\right)^{\mathrm{2}}…
Question Number 157436 by MathSh last updated on 23/Oct/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}\geqslant\mathrm{0}\:\:\mathrm{then}: \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}} \:+\:\mathrm{2}\:\geqslant\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{y}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{z}}} \\ $$$$ \\ $$ Answered by Jamshidbek last updated on…
Question Number 157438 by apriadodir last updated on 23/Oct/21 $$\mathrm{find}\:\mathrm{all}\:\mathrm{subgroups}\:\mathrm{of}\:: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{grup}\:\left(\mathrm{Z}_{\mathrm{6}} \:,\:+\right) \\ $$$$\left.\mathrm{b}\right)\:\mathrm{grup}\:\left(\mathrm{Z}_{\mathrm{6}} \:−\left\{\mathrm{0}\right\},\:×\right) \\ $$ Answered by mindispower last updated on 23/Oct/21…
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Question Number 157435 by MathSh last updated on 23/Oct/21 Answered by mindispower last updated on 23/Oct/21 $$\sim\frac{{ln}^{\Phi^{\mathrm{2}} −\varphi\Phi} \left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}},{x}\rightarrow\infty \\ $$$$\Phi^{\mathrm{2}} −\varphi\Phi>\mathrm{0}\:{not}\:{integrable}\:{sir}\:{tcheck}\:{it} \\ $$ Commented…
Question Number 26347 by tawa tawa last updated on 24/Dec/17 $$\mathrm{solve}: \\ $$$$\mathrm{5x}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} }\right)\:=\:\mathrm{12}\:\:\:\:\:\:\:\:…..\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{5y}\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} }\right)\:=\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:……\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$ Terms of Service Privacy…
Question Number 157417 by MathSh last updated on 23/Oct/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b}>\mathrm{0}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\left(\mathrm{a}\:+\:\mathrm{b}\right)^{\mathrm{6}} \:\left(\mathrm{a}^{\mathrm{15}} \:+\:\mathrm{b}^{\mathrm{15}} \right)\:\left(\mathrm{a}^{\mathrm{21}} \:+\:\mathrm{b}^{\mathrm{21}} \right)}{\left(\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \right)^{\mathrm{2}} \:\left(\mathrm{a}^{\mathrm{5}} \:+\:\mathrm{b}^{\mathrm{5}} \right)^{\mathrm{3}} \:\left(\mathrm{a}^{\mathrm{7}} \:+\:\mathrm{b}^{\mathrm{7}} \right)^{\mathrm{3}}…
Question Number 26348 by RoachDN last updated on 24/Dec/17 Commented by RoachDN last updated on 24/Dec/17 $$\mathrm{short}\:\mathrm{trick}\:\mathrm{to}\:\mathrm{solve}? \\ $$ Answered by $@ty@m last updated on…