Question Number 144877 by imjagoll last updated on 30/Jun/21 $$\:\:\mathrm{u}+\sqrt{\mathrm{u}}+\sqrt[{\mathrm{3}}]{\mathrm{u}}+\sqrt[{\mathrm{4}}]{\mathrm{u}}+\sqrt[{\mathrm{5}}]{\mathrm{u}}+…\:+\infty=? \\ $$$$ \\ $$ Answered by MJS_new last updated on 30/Jun/21 $$\forall{u}>\mathrm{0}:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{u}^{\mathrm{1}/{n}} \:=+\infty…
Question Number 79340 by TawaTawa last updated on 24/Jan/20 Commented by john santu last updated on 24/Jan/20 $${let}\:{this}\:{number}\::\:{a},{ar},{ar}^{\mathrm{2}} \\ $$$$\left({i}\right)\:{a}\left(\frac{{r}^{\mathrm{3}} −\mathrm{1}}{{r}−\mathrm{1}}\right)={p}\Rightarrow\frac{{r}^{\mathrm{3}} −\mathrm{1}}{{r}−\mathrm{1}}=\frac{{p}}{{a}} \\ $$$${r}^{\mathrm{2}} +{r}+\mathrm{1}=\frac{{p}}{{a}}\:…
Question Number 144860 by mathdanisur last updated on 29/Jun/21 $$\mid\frac{{x}}{{x}-\mathrm{1}}\mid\:+\:\mid{x}\mid\:=\:\frac{{x}^{\mathrm{2}} }{\mid{x}-\mathrm{1}\mid}\:\:\:{find}\:\:{x}=? \\ $$ Commented by hknkrc46 last updated on 29/Jun/21 $$\bigstar\:\mid\boldsymbol{{x}}\mid\:=\:\frac{\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mid\boldsymbol{{x}}\mid}{\mid\boldsymbol{{x}}\:−\:\mathrm{1}\mid} \\ $$$$\bigstar\:\mid\boldsymbol{{x}}^{\mathrm{2}} \:−\:\boldsymbol{{x}}\mid\:=\:\boldsymbol{{x}}^{\mathrm{2}}…
Question Number 144869 by mathdanisur last updated on 29/Jun/21 $${if}\:\:\mathrm{3}^{\boldsymbol{{z}}} \:=\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}\sqrt{\mathrm{3}}} \:\centerdot\:\mathrm{3}^{\mathrm{2}} }\:\:{find}\:\:\boldsymbol{{z}}=? \\ $$ Answered by liberty last updated on 30/Jun/21 $$\:\mathrm{3}^{\mathrm{z}} \:=\:\mathrm{3}^{−\left(\mathrm{2}+\mathrm{5}\sqrt{\mathrm{3}}\right)} \:\Rightarrow\mathrm{z}=−\mathrm{2}−\mathrm{5}\sqrt{\mathrm{3}}…
Question Number 144858 by mathdanisur last updated on 29/Jun/21 Answered by mindispower last updated on 29/Jun/21 $$\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\left({n}+\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} ={a}^{\mathrm{2}} \\ $$$$\left({n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\left(\left({n}−\mathrm{1}\right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 144839 by mathdanisur last updated on 29/Jun/21 $${If}\:\:{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{5}}\:+\:\mathrm{3}\:\:{and}\:\:{y}\:=\:\mathrm{4}\:\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$$${Prove}\:{that}:\:\:{x}\:-\:{y}\:<\:\mathrm{0} \\ $$ Answered by ajfour last updated on 29/Jun/21 $$\left({x}−\mathrm{3}\right)^{\mathrm{3}} =\mathrm{5} \\ $$$${y}^{\mathrm{3}}…
Question Number 79306 by jagoll last updated on 24/Jan/20 $$ \\ $$$$\sqrt{\mathrm{3}−\mathrm{x}}−\sqrt{\mathrm{x}+\mathrm{1}}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by kaivan.ahmadi last updated on 24/Jan/20 $$\mathrm{3}−{x}\geqslant\mathrm{0}\Rightarrow{x}\leqslant\mathrm{3} \\ $$$${x}+\mathrm{1}\geqslant\mathrm{0}\Rightarrow{x}\geqslant−\mathrm{1} \\…
Question Number 144831 by mathdanisur last updated on 29/Jun/21 $$\frac{\mathrm{3}}{\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}}\:+\:\frac{\mathrm{5}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}\:+\:\frac{\mathrm{7}}{\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}}\:+\:\frac{\mathrm{9}}{\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6}}\:+\:…\:\infty=? \\ $$ Answered by Dwaipayan Shikari last updated on 29/Jun/21 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}…
Question Number 144825 by mathdanisur last updated on 29/Jun/21 $$\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{k}}{{k}^{\mathrm{4}} \:+\:\mathrm{4}}\:=\:? \\ $$ Answered by Dwaipayan Shikari last updated on 29/Jun/21 $$\underset{{k}=\mathrm{1}} {\overset{\infty}…
Question Number 144826 by mathdanisur last updated on 29/Jun/21 Answered by mindispower last updated on 29/Jun/21 $${sin}^{−} \left({a}\right)+{sin}^{−} \left({b}\right)={u} \\ $$$${a},{b}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${sin}\left({u}\right)={a}\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }+{b}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }…