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} \\ $$$${a},{b}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${sin}\left({u}\right)={a}\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }+{b}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }…
Question Number 144820 by mathdanisur last updated on 29/Jun/21 Answered by mathmax by abdo last updated on 30/Jun/21 $$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}+\sqrt{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}\right)}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{x}=\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…
Question Number 144823 by loveineq last updated on 29/Jun/21 $$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+\mathrm{1}\:=\:\mathrm{3}{ab}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}+\mathrm{1}}{{b}+\mathrm{1}}+\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\:\leqslant\:{a}+{b} \\ $$ Answered by ArielVyny last updated on 30/Jun/21 $${we}\:{know}\:{that}\:{a}+{b}+\mathrm{1}=\mathrm{3}{ab} \\ $$$${we}\:{suppose}\:{ab}\geqslant\mathrm{1} \\…
Question Number 79279 by jagoll last updated on 24/Jan/20 $$\mathrm{solve}\: \\ $$$$\mid\mathrm{x}\mid^{\mathrm{3}} −\mathrm{7x}^{\mathrm{2}} +\mathrm{7}\mid\mathrm{x}\mid+\mathrm{15}<\mathrm{0} \\ $$ Commented by jagoll last updated on 24/Jan/20 $$\mathrm{thanks}\: \\…
Question Number 13731 by prakash jain last updated on 22/May/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{p}>{q}>\mathrm{0}\:\mathrm{and}\:{x}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{p}}\left(\frac{{x}^{{p}} }{{p}+\mathrm{1}}−\mathrm{1}\right)\geqslant\frac{\mathrm{1}}{{q}}\left(\frac{{x}^{{q}} }{{q}+\mathrm{1}}−\mathrm{1}\right).\: \\ $$ Commented by mrW1 last updated on 25/May/17 $${function}\:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\left(\frac{{a}^{{x}}…