Question Number 161123 by HongKing last updated on 12/Dec/21 $$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{H}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}\left(\mathrm{H}_{\mathrm{2}\boldsymbol{\mathrm{n}}-\mathrm{1}} \:-\:\mathrm{2}\:\mathrm{H}_{\boldsymbol{\mathrm{n}}-\mathrm{1}} \right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161126 by geron last updated on 12/Dec/21 Commented by geron last updated on 12/Dec/21 $$\mathrm{79}\:? \\ $$ Commented by geron last updated on…
Question Number 161096 by HongKing last updated on 11/Dec/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{different}\:\mathrm{in}\:\mathrm{pairs}\:\mathrm{and}\:\:\mathrm{n};\mathrm{k}\in\mathbb{N}^{\ast} \\ $$$$\frac{\mathrm{log}\:\mathrm{x}^{\boldsymbol{\mathrm{n}}} }{\mathrm{b}^{\boldsymbol{\mathrm{k}}} \:-\:\mathrm{c}^{\boldsymbol{\mathrm{k}}} }\:=\:\frac{\mathrm{log}\:\mathrm{y}^{\boldsymbol{\mathrm{n}}} }{\mathrm{c}^{\boldsymbol{\mathrm{k}}} \:-\:\mathrm{a}^{\boldsymbol{\mathrm{k}}} }\:=\:\frac{\mathrm{log}\:\mathrm{z}^{\boldsymbol{\mathrm{n}}} }{\mathrm{a}^{\boldsymbol{\mathrm{k}}} \:-\:\mathrm{b}^{\boldsymbol{\mathrm{k}}} } \\ $$$$\mathrm{then}\:\mathrm{find}\:\:\sqrt{\boldsymbol{\mathrm{xyz}}}…
Question Number 161102 by cortano last updated on 12/Dec/21 Commented by Abdulazizov last updated on 12/Dec/21 $$ \\ $$ Answered by mr W last updated…
Question Number 95560 by I want to learn more last updated on 26/May/20 Commented by i jagooll last updated on 26/May/20 $$\mathrm{y}=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{5}\right)^{−\mathrm{1}} \: \\…
Question Number 161091 by HongKing last updated on 11/Dec/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\mathrm{1}\:-\:\mathrm{x}}\:=\:\mathrm{2x}^{\mathrm{2}} \:-\:\mathrm{1}\:-\:\mathrm{2x}\:\sqrt{\mathrm{1}\:-\:\mathrm{x}^{\mathrm{2}} } \\ $$$$ \\ $$ Answered by MJS_new last updated on 11/Dec/21…
Question Number 161075 by mnjuly1970 last updated on 11/Dec/21 $$ \\ $$$$\:\:\:\:{simplify} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{n}}{\left(\:{n}^{\:\mathrm{2}} −\frac{\:\mathrm{1}}{\mathrm{4}}\:\right)^{\:\mathrm{3}} }\:=\:? \\ $$$$ \\ $$ Terms…
Question Number 161068 by blackmamba last updated on 11/Dec/21 $$\:\:\:\:\:\int\:\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}\:{dx}\:=? \\ $$ Answered by chhaythean last updated on 11/Dec/21 $$=−\int\frac{\mathrm{2x}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{4}} −\mathrm{1}}}\mathrm{dx} \\…
Question Number 161065 by cortano last updated on 11/Dec/21 $$\:\begin{cases}{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:+\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{a}}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{b}}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{4}}]{{x}−{abc}}\:=\:{c}}\end{cases} \\ $$$$\:{find}\:\sqrt{{x}+{abc}}\:+\sqrt{{x}−{abc}} \\ $$ Answered by Rasheed.Sindhi last updated on 12/Dec/21 $$\:\begin{cases}{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:+\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{a}….\left({i}\right)}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{b}…..\left({ii}\right)}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{4}}]{{x}−{abc}}\:=\:{c}…..\left({iii}\right)}\end{cases} \\ $$$$\mathrm{2}\sqrt[{\mathrm{4}}]{{x}+{abc}}\:={a}+{b} \\…
Question Number 161059 by HongKing last updated on 11/Dec/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}\right)\mathrm{dxdy} \\ $$ Terms of Service Privacy Policy Contact:…