Question Number 87817 by M±th+et£s last updated on 06/Apr/20 $${f}\left(\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}\right)+{f}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{1}}\right)={x} \\ $$$${find}\:{f}\left({x}\right) \\ $$ Commented by M±th+et£s last updated on 06/Apr/20 $${like}\:\mathrm{87755}\:{question}\:{but}\:{here}\:\left({x}−\mathrm{1}\right) \\ $$ Commented…
Question Number 153339 by amin96 last updated on 06/Sep/21 $$\:\:\frac{{x}−\mathrm{1}}{{x}}+\frac{{x}−\mathrm{2}}{{x}}+\frac{{x}−\mathrm{3}}{{x}}+\ldots+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:\:\:\:{x}=? \\ $$$$\because\therefore\because\therefore\because\:\:{Easy}\:{question}\therefore\because\therefore\because\therefore\because \\ $$ Answered by MJS_new last updated on 06/Sep/21 $$\frac{\underset{{j}=\mathrm{1}} {\overset{{x}−\mathrm{1}} {\sum}}{j}}{{x}}=\frac{{x}−\mathrm{1}}{\mathrm{2}}=\mathrm{3}\:\Rightarrow\:{x}=\mathrm{7} \\…
Question Number 22244 by ajfour last updated on 14/Oct/17 $${Solve}\:{the}\:{inequality}\:: \\ $$$$\:−\mathrm{9}\left(\sqrt[{\mathrm{4}}]{{x}}\right)+\sqrt{{x}}+\mathrm{18}\:\geqslant\:\mathrm{0}\:. \\ $$ Answered by $@ty@m last updated on 14/Oct/17 $${Let}\:\sqrt{{x}}={y} \\ $$$$−\mathrm{9}\sqrt{{y}}+{y}+\mathrm{18}\geqslant\mathrm{0} \\…
Question Number 22220 by Tinkutara last updated on 13/Oct/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{4log}_{{x}/\mathrm{2}} \left(\sqrt{{x}}\right)+\mathrm{2log}_{\mathrm{4}{x}} \left({x}^{\mathrm{2}} \right)= \\ $$$$\mathrm{3log}_{\mathrm{2}{x}} \left({x}^{\mathrm{3}} \right)\:\mathrm{is} \\ $$ Answered by ajfour last…
Question Number 87755 by john santu last updated on 06/Apr/20 $$\mathrm{f}\left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{1}}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\mathrm{x} \\ $$$$\mathrm{find}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$ Commented by john santu last updated on 06/Apr/20 $$\left(\mathrm{i}\right)\:\mathrm{let}\:\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{1}}\:=\:\mathrm{x}'\:\Rightarrow\mathrm{f}\left(\mathrm{x}'\right)\:+\:\mathrm{f}\left(\frac{\mathrm{x}'−\mathrm{3}}{\mathrm{x}'+\mathrm{1}}\right)\:=\:\frac{\mathrm{x}'+\mathrm{3}}{\mathrm{1}−\mathrm{x}} \\…
Question Number 22199 by chernoaguero@gmail.com last updated on 13/Oct/17 Commented by chernoaguero@gmail.com last updated on 13/Oct/17 $$\mathrm{Thank}\:\mathrm{you}\:\mathrm{guys} \\ $$ Commented by Rasheed.Sindhi last updated on…
Question Number 87737 by M±th+et£s last updated on 05/Apr/20 $${solve} \\ $$$${sin}\left(\frac{\pi}{\left[\frac{\left[{x}\right]}{\mathrm{4}}\right]}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by mahdi last updated on 06/Apr/20 $$\mathrm{u}=\left[\frac{\left[\mathrm{x}\right]}{\mathrm{4}}\right]\Rightarrow−\mathrm{1}\leqslant\frac{\mathrm{1}}{\mathrm{u}}\leqslant\mathrm{1}\Rightarrow−\pi\leqslant\frac{\pi}{\mathrm{u}}\leqslant\pi\:\:\:\left\{\mathrm{u}\neq\mathrm{0}\Rightarrow\mathrm{x}\notin\left[\mathrm{0},\mathrm{4}\right)\right\} \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{u}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\begin{cases}{\frac{\pi}{\mathrm{u}}=\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi}\\{\frac{\pi}{\mathrm{u}}=\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2k}\pi}\end{cases} \\…
Question Number 87733 by TawaTawa1 last updated on 05/Apr/20 Commented by mahdi last updated on 05/Apr/20 $$\mathrm{27y}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }=\left(\mathrm{3y}+\frac{\mathrm{1}}{\mathrm{y}}\right)\left(\mathrm{9y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }−\mathrm{3}\right)= \\ $$$$\left(\mathrm{3y}+\frac{\mathrm{1}}{\mathrm{y}}\right)\left(\mathrm{3}−\mathrm{3}\right)=\mathrm{0} \\ $$…
Question Number 87726 by A8;15: last updated on 05/Apr/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 87724 by mr W last updated on 05/Apr/20 $${solve}\:{the}\:{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{cos}}\:\lfloor\boldsymbol{{x}}\rfloor\right)=\mathrm{1} \\ $$ Answered by mahdi last updated on 05/Apr/20 $$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\left[\mathrm{x}\right]\right)=\mathrm{1}\Rightarrow\mathrm{cos}\left[\mathrm{x}\right]=\mathrm{sin}\left(\mathrm{1}+\mathrm{2k}\pi\right)…