Question Number 81913 by ahmadshahhimat775@gmail.com last updated on 16/Feb/20 Commented by john santu last updated on 16/Feb/20 $$\Rightarrow\sqrt{{x}}+\sqrt[{\mathrm{6}\:}]{{y}}\:+\frac{{y}\:\sqrt[{\mathrm{6}\:}]{{y}}+{x}\:\sqrt[{\mathrm{6}\:}]{{x}}}{\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt{{y}}}\:= \\ $$$$\frac{\sqrt[{\mathrm{6}\:}]{{x}^{\mathrm{5}} \:}\sqrt{{y}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt[{\mathrm{6}\:}]{{y}^{\mathrm{4}} }+\sqrt[{\mathrm{6}\:}]{{y}^{\mathrm{7}} }+\sqrt[{\mathrm{6}\:}]{{x}^{\mathrm{7}} }}{\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt{{y}}} \\…
Question Number 16374 by gourav~ last updated on 21/Jun/17 $${The}\:{Value}\:{of}\:{the}\:{sum}..\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right),\:{Where}\:{i}=\sqrt{−\mathrm{1}\:\:} \\ $$$${is}.. \\ $$$$\left({a}.\right)\:{i} \\ $$$$\left({b}.\right)\:{i}−\mathrm{1} \\ $$$$\left({c}.\right)\:−{i} \\ $$$$\left({d}.\right)\:\mathrm{0} \\…
Question Number 16373 by gourav~ last updated on 21/Jun/17 $${if}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{200}} {\sum}}{i}^{{k}} +\underset{{p}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}{i}^{{p}} ={x}+{iy}\:{then}..\left({x},{y}\right){is}… \\ $$$${a}.\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${b}.\:\left(\mathrm{1},−\mathrm{1}\right) \\ $$$${c}.\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${d}.\:\left(\mathrm{4},\mathrm{8}\right) \\…
Question Number 147432 by mathdanisur last updated on 20/Jul/21 $${if}\:\:\begin{cases}{\mathrm{2}{x}\:+\:{a}\:\:;\:\:{x}\:<\:−\mathrm{3}}\\{{x}^{\mathrm{2}} \:-\:\mathrm{4}\:\:;\:\:−\mathrm{3}\:\leqslant\:{x}\:<\:\mathrm{2}}\\{{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:\:;\:\:{x}\:\geqslant\:\mathrm{2}}\end{cases} \\ $$$${find}\:\:\:\mathrm{3}{a}\:-\:{b}\:=\:? \\ $$ Answered by liberty last updated on 21/Jul/21 $$\left(\bullet\right)\:\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow−\mathrm{3}}…
Question Number 147431 by vvvv last updated on 20/Jul/21 Answered by SEKRET last updated on 21/Jul/21 $$\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{c}}=\boldsymbol{\mathrm{d}}=\boldsymbol{\mathrm{x}} \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{x}}_{\mathrm{1}} \centerdot\boldsymbol{\mathrm{x}}_{\mathrm{2}} \centerdot\boldsymbol{\mathrm{x}}_{\mathrm{3}} \centerdot\boldsymbol{\mathrm{x}}_{\mathrm{4}} =\:\boldsymbol{\mathrm{w}}=\:\boldsymbol{\mathrm{abcd}}…
Question Number 81892 by Power last updated on 16/Feb/20 Answered by MJS last updated on 16/Feb/20 $$\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{any}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{next}\:\mathrm{number}\:\Rightarrow\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{not}\:\mathrm{unique} \\ $$ Commented by mr W…
Question Number 81871 by mr W last updated on 16/Feb/20 $${a}_{\mathrm{1}} =\mathrm{4} \\ $$$${a}_{{n}+\mathrm{1}} =\frac{\mathrm{4}{a}_{{n}} +\mathrm{3}}{{a}_{{n}} +\mathrm{2}} \\ $$$${find}\:{a}_{{n}} =? \\ $$ Commented by jagoll…
Question Number 147402 by vvvv last updated on 20/Jul/21 $$\boldsymbol{{f}}\left(\boldsymbol{{x}}+\mathrm{1}\right)\left(\boldsymbol{{x}}+\mathrm{2}\right)….\left(\boldsymbol{{x}}+\mathrm{2018}\right) \\ $$$$\boldsymbol{{f}}'\left(\mathrm{0}\right)=? \\ $$ Answered by SEKRET last updated on 21/Jul/21 $$\:\:\boldsymbol{\mathrm{y}}=\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)\centerdot….\left(\boldsymbol{\mathrm{x}}+\mathrm{2018}\right) \\ $$$$\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{y}}\right)\:=\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)+\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)+..+\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{2018}\right) \\…
Question Number 147364 by henderson last updated on 20/Jul/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{T}}\:=\:\boldsymbol{{x}}^{\mathrm{5}} +\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{reductible}}\:\boldsymbol{\mathrm{in}}\:\mathbb{Q}\:? \\ $$ Answered by Olaf_Thorendsen last updated on 20/Jul/21 $$\mathrm{T}\in\mathbb{Q}\left[\mathrm{X}\right],\:\mathrm{T}\left({x}\right)\:=\:{x}^{\mathrm{5}} +\mathrm{3}{x}^{\mathrm{2}}…
Question Number 147371 by mathdanisur last updated on 20/Jul/21 $${if}\:\:{x}>\mathrm{0}\:\:{then}: \\ $$$${e}^{\boldsymbol{{x}}+\boldsymbol{{e}}^{−\boldsymbol{{x}}} } \:+\:{e}^{−\boldsymbol{{x}}+\boldsymbol{{e}}^{\boldsymbol{{x}}} } \:\geqslant\:\mathrm{2}{coshx}\:\centerdot\:{e}^{\boldsymbol{{sechx}}} \\ $$ Answered by mindispower last updated on 20/Jul/21…