Question Number 100943 by Ar Brandon last updated on 29/Jun/20 $$\mathcal{D}\mathrm{etermine}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}; \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{5}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} −\mathrm{1}} \\ $$ Answered by mathmax by abdo last updated on…
Question Number 35384 by 7991 last updated on 18/May/18 $$\sqrt[{\mathrm{6}}]{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}\:=….?? \\ $$ Commented by prof Abdo imad last updated on 18/May/18 $${the}\:\mathrm{6}^{{eme}} \:{roots}\:{of}\:−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\:{are}\:{the}\:{complex}\:{z}_{{k}} \\ $$$${wich}\:{verify}\:{z}_{{k}}…
Question Number 100916 by 175 last updated on 29/Jun/20 $${solve}\:{the}\:{eqution}\:: \\ $$$$\frac{\mathrm{2}\:+\:{x}}{\mathrm{12}\:+\:\mathrm{4}{x}}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \:\:\:\:\:\:\:.,{x}\:=\mathrm{2}\: \\ $$ Answered by 1549442205 last updated on 30/Jun/20 $$\Leftrightarrow\mathrm{2}^{\mathrm{x}} −\frac{\mathrm{12}+\mathrm{4x}}{\mathrm{2}+\mathrm{x}}=\mathrm{0}\Leftrightarrow\mathrm{2}^{\mathrm{x}} −\frac{\mathrm{6}}{\mathrm{x}+\mathrm{2}}−\mathrm{4}=\mathrm{0}\left(\ast\right)…
Question Number 100904 by Dwaipayan Shikari last updated on 29/Jun/20 $$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\left[\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)……\mathrm{3}{n}}{{n}^{\mathrm{2}{n}} }\right]^{\frac{\mathrm{1}}{{n}}} \\ $$ Commented by Ar Brandon last updated on 29/Jun/20 I like how you transformed the sum directly into integral without decomposing it into simpler sums.�� Commented…
Question Number 35367 by Rio Mike last updated on 18/May/18 $${Q}_{\mathrm{3}\:} .\:{a}\:{number}\:{is}\:{a}\:{factor}\:{of}\:\mathrm{6} \\ $$$$\:{the}\:{square}\:{of}\:{the}\:{number}\:{added}\: \\ $$$${to}\:\mathrm{5}\:{times}\:{thenumber}\:{is}\:{same}\:{as} \\ $$$$−\mathrm{6}\:{find}\:{the}\:{number}. \\ $$ Answered by MJS last updated…
Question Number 35363 by Rio Mike last updated on 18/May/18 $${Q}_{\mathrm{1}} .\:{A}\:{quadratic}\:{equation}\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}=\mathrm{0} \\ $$$${has}\:{roots}\:\alpha\:{and}\:\beta\:.{without}\:{solving} \\ $$$$\left.{a}\right){write}\:{down}\:{the}\:{values}\:{of}\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \\ $$$$\left.{b}\right)\:{find}\:{the}\:{quadratic}\:{equation} \\ $$$${with}\:{integral}\:{coefficients},{whose} \\ $$$${roots}\:{are}\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:{and}\frac{\mathrm{1}}{\beta^{\mathrm{2}}…
Question Number 35357 by Rio Mike last updated on 18/May/18 $${Q}_{\mathrm{1}} .\:{find}\:{the}\:{term}\:{in}\:{x}^{\mathrm{6}} \:{in}\:{the}\: \\ $$$${expansion}\:{of}\:\left({x}^{\mathrm{2}} +\frac{\mathrm{2}}{{x}}\right)^{\mathrm{9}} \\ $$$${Q}_{\mathrm{2}} .\:{in}\:{the}\:{binomial}\:{expansion}\:{of} \\ $$$$\:\left({x}+\frac{{k}}{{x}}\right)^{\mathrm{6}} ,\:{the}\:{term}\:{independent}\:{of}\:{x} \\ $$$${is}\:\mathrm{160}\:{find}\:{the}\:{value}\:{of}\:{k}. \\…
Question Number 35304 by Rio Mike last updated on 17/May/18 $$\left.\:{Q}\mathrm{1}.\:\:\:{a}\right)\:{solve}\:{for}\:{x}\:\:\mathrm{9}^{{x}} +\mathrm{5}\left(\mathrm{3}^{{x}} \right)=\mathrm{6} \\ $$$$\left.{b}\right){write}\:{down}\:{the}\:{first}\:\:\mathrm{4}\:{terms} \\ $$$${in}\:{the}\:{binomial}\:{expansion}\:{of}\:\left(\mathrm{1}−\mathrm{3}{x}\right)^{\mathrm{7}} \\ $$$$\left.{c}\right){the}\:{sum}\:{S}_{{n}} \:{of}\:{the}\:{first}\:{n}^{{th}} {terms} \\ $$$${is}\:{given}\:{by}\:{S}_{{n}\:} =\:\mathrm{3}\left(\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \right)\:{find}…
Question Number 35297 by 123 45 polytechnicien last updated on 17/May/18 $${sokve}\:{x}^{\mathrm{3}} +\mathrm{6}{y}^{\mathrm{3}} =\mathrm{4}{z}^{\mathrm{3}} \:{x}\:{y}\:{z}\:{integers} \\ $$ Commented by Rasheed.Sindhi last updated on 18/May/18 $$\mathrm{4z}^{\mathrm{3}}…
Question Number 166346 by LEKOUMA last updated on 18/Feb/22 $$\int\frac{{dx}}{\mathrm{1}+\sqrt{{x}}+\sqrt{\mathrm{1}+{x}}} \\ $$ Commented by mkam last updated on 18/Feb/22 $$\boldsymbol{{w}}\:=\:\mathrm{1}\:+\:\sqrt{\boldsymbol{{x}}}\:+\sqrt{\mathrm{1}+\boldsymbol{{x}}}\:\Rightarrow\:\boldsymbol{{dw}}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\boldsymbol{{x}}}\:+\mathrm{2}\sqrt{\mathrm{1}+\boldsymbol{{x}}}}\right)\boldsymbol{{dx}} \\ $$$$ \\ $$$$\Rightarrow\:\boldsymbol{{dw}}\:=\:\left(\:\frac{\mathrm{1}}{\mathrm{2}\:\boldsymbol{{w}}\:−\:\mathrm{2}}\:\right)\:\boldsymbol{{dx}}\:\Rightarrow\:\boldsymbol{{dw}}\:\left(\:\mathrm{2}\:\boldsymbol{{w}}\:−\:\mathrm{2}\:\right)\:=\:\boldsymbol{{dx}} \\…