Question Number 100660 by bobhans last updated on 28/Jun/20 $$\mid{x}^{\mathrm{2}} −{x}\mid\:<\:\mathrm{2}+{x}\:.\:{find}\:{solution}\:{set}. \\ $$ Commented by bramlex last updated on 28/Jun/20 $${since}\:\mid{x}^{\mathrm{2}} −{x}\mid\:\geqslant\mathrm{0}\:{then}\:\mathrm{2}+{x}\:{must}\:{be}\:>\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}+{x}\:>\:\mathrm{0}\:;\:{x}\:>−\mathrm{2} \\…
Question Number 166195 by mathlove last updated on 15/Feb/22 Answered by alephzero last updated on 15/Feb/22 $${x}\:=\:\mathrm{1} \\ $$ Commented by mathlove last updated on…
Question Number 166181 by amin96 last updated on 14/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166182 by SANOGO last updated on 14/Feb/22 $${calculer}\:{la}\:{primitive}\:{de} \\ $$$$\int\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$ Answered by greogoury55 last updated on 14/Feb/22 $${t}=\mathrm{tan}\:{u}…
Question Number 166160 by cortano1 last updated on 14/Feb/22 $$\:\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\mathrm{m}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{m}^{\mathrm{2}} \mathrm{n}+\mathrm{mn}^{\mathrm{2}} +\mathrm{2mn}}=? \\ $$ Answered by Jamshidbek last updated on 14/Feb/22…
Question Number 100622 by MJS last updated on 27/Jun/20 Commented by MJS last updated on 27/Jun/20 $$\mathrm{building}\:\mathrm{on}\:\mathrm{my}\:\mathrm{work},\:\mathrm{can}\:\mathrm{someone}\:\mathrm{try}\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{for}\:{x}\in\mathbb{Q}\:\mathrm{and} \\ $$$${x}\left[{x}\left[{x}\left[{x}\right]\right]\right]={n}\in\mathbb{N}\:? \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{picture},\:{i}\:\mathrm{and}\:{f}\:\mathrm{are}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:{x}…
Question Number 35080 by math1967 last updated on 15/May/18 $${If}\:\frac{{a}}{{b}+{c}}\:+\frac{{b}}{{c}+{a}}\:+\frac{{c}}{{a}+{b}}=\mathrm{1}\:{then}\:{prove}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{{b}+{c}}\:+\frac{{b}^{\mathrm{2}} }{{c}+{a}}\:+\frac{{c}^{\mathrm{2}} }{{a}+{b}}=\mathrm{0} \\ $$ Answered by ajfour last updated on 15/May/18 $$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}=\mathrm{1}…
Question Number 100597 by bobhans last updated on 27/Jun/20 Commented by Dwaipayan Shikari last updated on 27/Jun/20 $${x}+\mathrm{2}\mid{y}\mid=\mathrm{3} \\ $$$${First}\:{case}\:\:\:\:{x}+\mathrm{2}{y}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}−\mathrm{3}{y}=\mathrm{5}\:\:\:\Rightarrow\mathrm{5}{y}=−\mathrm{2}\:\:\Rightarrow{y}=−\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:{x}=\frac{\mathrm{19}}{\mathrm{5}} \\ $$$${x}−{y}=\frac{\mathrm{21}}{\mathrm{5}}\:\:\:\:\:\:\begin{cases}{{x}=\frac{\mathrm{19}}{\mathrm{5}}}\\{{y}=−\frac{\mathrm{2}}{\mathrm{5}}}\end{cases} \\…
Question Number 35056 by math khazana by abdo last updated on 14/May/18 $${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{jx}\right)^{{n}} \:−\left(\mathrm{1}−{jx}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){factorize}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$${j}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:. \\ $$ Commented by…
Question Number 166125 by mathlove last updated on 13/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com