Question Number 213486 by a.lgnaoui last updated on 06/Nov/24 $$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\mathrm{5}\boldsymbol{\mathrm{x}}−\frac{\mathrm{6}}{\boldsymbol{\mathrm{x}}}=\mathrm{0}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}? \\ $$ Answered by Frix last updated on 06/Nov/24 $${x}^{\mathrm{6}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{6}}…
Question Number 213482 by efronzo1 last updated on 06/Nov/24 Answered by A5T last updated on 06/Nov/24 $$\mathrm{3}\mid{Y}\Rightarrow\mathrm{8}+{c}+\mathrm{2}+\mathrm{3}+{d}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow{c}+{d}\equiv\mathrm{2}\left({mod}\:\mathrm{3}\right) \\ $$$$\mathrm{3}\mid{X}\Rightarrow\mathrm{9}\mid{Y}\::\: \\ $$$$\mathrm{3}+\mathrm{1}+{a}+\mathrm{5}+{b}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow\mathrm{8}+{c}+\mathrm{2}+\mathrm{3}+{d}\equiv\mathrm{0}\left({mod}\:\mathrm{9}\right) \\ $$$$\Rightarrow{a}+{b}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow{c}+{d}\equiv\mathrm{5}\left({mod}\:\mathrm{9}\right) \\ $$$$\mathrm{3}\mid{X}\Rightarrow{c}+{d}=\mathrm{5}\Rightarrow{a}+{b}=\mathrm{0},\mathrm{3},\mathrm{6},\mathrm{9}\:\wedge\:{c}+{d}=\mathrm{5}…
Question Number 213439 by Ari last updated on 05/Nov/24 Commented by BaliramKumar last updated on 06/Nov/24 $${l}\:{and}\:{b}\:=\:? \\ $$$$\mathrm{If}\:\mathrm{square}\:\:\:\:=\:\lceil\frac{\sqrt{\mathrm{2000}}}{\mathrm{7}}\rceil^{\mathrm{2}} =\:\mathrm{49} \\ $$$$ \\ $$ Commented…
Question Number 213430 by hardmath last updated on 05/Nov/24 $$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{x}\:+\:\left[\mathrm{y}\right]\:+\:\left\{\mathrm{z}\right\}\:=\:\mathrm{9},\mathrm{4}}\\{\left[\mathrm{x}\right]\:+\:\left\{\mathrm{y}\right\}\:+\:\mathrm{z}\:=\:\mathrm{11},\mathrm{3}}\\{\left\{\mathrm{x}\right\}\:+\:\mathrm{y}\:+\:\left[\mathrm{z}\right]\:=\:\mathrm{10},\mathrm{5}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by A5T last updated on 05/Nov/24 $$\left({i}\right)−\left({ii}\right):\:\left\{{x}\right\}+\left[{y}\right]−\left\{{y}\right\}−\left[{z}\right]=−\mathrm{1}.\mathrm{9}…\left({iv}\right) \\ $$$$\left({iv}\right)−\left({iii}\right)\Rightarrow\left[{y}\right]−\left\{{y}\right\}−{y}−\mathrm{2}\left[{z}\right]=−\mathrm{12}.\mathrm{4} \\…
Question Number 213417 by hardmath last updated on 04/Nov/24 $$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}\:,\:\mathrm{d}\:\in\:\mathbb{N} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:+\:\mathrm{d}\:=\:\mathrm{63} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{maksimum}\left(\mathrm{ab}\:+\:\mathrm{bc}\:+\:\mathrm{cd}\right)\:=\:? \\ $$ Answered by Frix last updated on 05/Nov/24 $${d}=\mathrm{63}−{a}−{b}−{c} \\…
Question Number 213391 by York12 last updated on 04/Nov/24 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{where}\:{a},{b},{c}\geqslant\mathrm{0} \\ $$$${a}−\mathrm{2}{bc}={b}−\mathrm{2}{ac}={c}−\mathrm{2}{ab} \\ $$$${a}+{b}+{c}=\mathrm{2}\: \\ $$ Answered by Frix last updated on 04/Nov/24 $$\mathrm{Due}\:\mathrm{to}\:\mathrm{symmetry}\:{a}={b}={c}=\frac{\mathrm{2}}{\mathrm{3}} \\…
Question Number 213413 by hardmath last updated on 04/Nov/24 $$\mathrm{Find}: \\ $$$$\mathrm{A}=\:\left[\sqrt{\mathrm{1}}\right]\:+\:\left[\sqrt{\mathrm{2}}\right]\:+\:\left[\sqrt{\mathrm{3}}\:\right]+…+\:\left[\sqrt{\mathrm{323}}\right]\:=\:? \\ $$ Answered by mehdee7396 last updated on 04/Nov/24 $$=\left(\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+\right)+\left(\left[\mathrm{4}\right]+\left[\mathrm{5}\right]+…+\left[\left[\sqrt{\mathrm{8}}\right]\right)\right. \\ $$$$+\left(\left[\sqrt{\mathrm{9}}\right]+\left[\sqrt{\mathrm{10}}\right]+…+\left[\sqrt{\mathrm{24}}\right]\right)+… \\…
Question Number 213374 by hardmath last updated on 03/Nov/24 $$\mathrm{Find}:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\sqrt{\mathrm{5}}}{\left(\sqrt{\mathrm{3}}\:−\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\right)^{\mathrm{2}} }\:=\:? \\ $$ Commented by Frix last updated on 03/Nov/24 $$\frac{\mathrm{29}−\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{8}−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{3}}{\mathrm{4}}} +\frac{\mathrm{13}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{\mathrm{18}−\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$…
Question Number 213264 by efronzo1 last updated on 02/Nov/24 $$\:\:\:\:\:\:\sqrt{\mathrm{2}}\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}\:+\:\mathrm{1}\:=\:\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\mathrm{x}_{\mathrm{1}} \:,\:\mathrm{x}_{\mathrm{2}} \:\:\underline{\underbrace{\square}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 213099 by MathematicalUser2357 last updated on 30/Oct/24 $$\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}\right)−\left({x}−\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${x}=… \\ $$ Answered by MrGaster last updated on 30/Oct/24 $$\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}−{x}+\mathrm{3}=\frac{\mathrm{1}}{\mathrm{3}}{x}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}}{x}+\frac{\mathrm{7}}{\mathrm{6}} \\…