Question Number 61343 by bhanukumarb2@gmail.com last updated on 01/Jun/19 Answered by MJS last updated on 01/Jun/19 $${x}={a}−\frac{\mathrm{1}}{{a}−\frac{\mathrm{1}}{{a}−…}} \\ $$$${x}={a}−\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} −{ax}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}…
Question Number 192409 by Abdullahrussell last updated on 17/May/23 Answered by Frix last updated on 17/May/23 $$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{2}} −\mathrm{3}{xyz} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\left({x}+{y}+{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}}…
Question Number 192399 by mnjuly1970 last updated on 17/May/23 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Algebra}\:\left(\mathrm{1}\:\right) \\ $$$$\:\:{G},\:{is}\:{a}\:{group}\:\:{and}\:\:\:{o}\left({G}\:\right)\:=\:{p}^{\:\mathrm{2}} \:. \\ $$$$\:\:\:{prove}\:{that}\:{G}\:{is}\:{an}\:{abelian}\:{group}. \\ $$$$\:\:\:{hint}:\:\:\left(\:{p}\:{is}\:{prime}\:{number}\:\:\right) \\ $$$$\:\:\:\:\:−−−−−−−−−−−−− \\ $$ Commented by aleks041103 last…
Question Number 61320 by behi83417@gmail.com last updated on 31/May/19 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}}: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\boldsymbol{\mathrm{x}}=\mathrm{12} \\ $$ Answered by MJS last updated on 31/May/19 $$\mathrm{no}\:“\mathrm{beautiful}''\:\mathrm{solution} \\ $$$${x}_{\mathrm{1}}…
Question Number 192387 by cortano12 last updated on 16/May/23 $$\:\mathrm{Simplify}\: \\ $$$$\:\sqrt{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{4}+\left(\frac{\mathrm{2017}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }\right)^{\mathrm{2}} }\right)}\: \\ $$$$\:\mathrm{is}\:….\: \\ $$ Answered by a.lgnaoui last updated on…
Question Number 126845 by mnjuly1970 last updated on 24/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:…\:{calculus}\:\:\left(\mathrm{I}\right)… \\ $$$$\:\:\:\:{prove}\:::\: \\ $$$$\:\:\:\:\:\:\:{i}::\:\:\lfloor\mathrm{2}{x}\rfloor\overset{?} {=}\lfloor{x}\rfloor+\lfloor{x}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor \\ $$$$\:\:\:\:\:\:\:{ii}::\:\lfloor\mathrm{3}{x}\rfloor=\lfloor{x}\rfloor+\lfloor{x}+\frac{\mathrm{1}}{\mathrm{3}}\rfloor+\lfloor{x}+\frac{\mathrm{2}}{\mathrm{3}}\rfloor \\ $$$$ \\ $$ Answered by floor(10²Eta[1]) last…
Question Number 192374 by mehdee42 last updated on 15/May/23 $${why}\:\:\:“\:\mathrm{200}!<\mathrm{100}^{\mathrm{200}} \:''\:? \\ $$ Commented by mehdee42 last updated on 17/May/23 $$\mathrm{99}×\mathrm{101}<\mathrm{100}^{\mathrm{2}} \\ $$$$\mathrm{98}×\mathrm{102}<\mathrm{100}^{\mathrm{2}} \\ $$$$:…
Question Number 192370 by Tawa11 last updated on 15/May/23 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\:\:\mathrm{x}^{\mathrm{4}} \:\:−\:\:\mathrm{2x}^{\mathrm{3}} \:\:+\:\:\mathrm{4x}^{\mathrm{2}} \:\:+\:\:\mathrm{6x}\:\:\:−\:\:\mathrm{21}\:\:\:=\:\:\:\mathrm{0},\:\: \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{of}\:\mathrm{its}\:\mathrm{roots}\:\mathrm{is}\:\mathrm{zero} \\ $$ Answered by Frix last updated on 16/May/23 $$\left({x}−{a}\right)\left({x}+{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)=\mathrm{0}…
Question Number 192363 by Red1ight last updated on 15/May/23 $$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}_{{i}} −{x}+\left(\mathrm{2}{cx}−{cb}\right)\left({y}_{{i}} +{cx}^{\mathrm{2}} −{cbx}\right)=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{this}\:\mathrm{equaition} \\ $$$$\left.{i}\right)\mathrm{c}>\mathrm{0} \\ $$$$\left.{ii}\right)\mathrm{b}>\mathrm{0} \\ $$$$\left.{iii}\right)\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution} \\ $$…
Question Number 192350 by Abdullahrussell last updated on 15/May/23 Answered by Frix last updated on 15/May/23 $$\mathrm{Just}\:\mathrm{type}\:\mathrm{into}\:\mathrm{a}\:\mathrm{calculator}. \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{5}}{\mathrm{197}} \\ $$$$\mathrm{But}\:\mathrm{can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{this}? \\ $$$$\frac{\left(\left({x}−\mathrm{5}\right)^{\mathrm{4}} +\mathrm{4}\right)\left(\left({x}−\mathrm{1}\right)^{\mathrm{4}} +\mathrm{4}\right)\left(\left({x}+\mathrm{3}\right)^{\mathrm{4}}…