Question Number 30092 by math1967 last updated on 16/Feb/18 $${If}\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:{find}\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} } \\ $$ Answered by math1967 last updated on 16/Feb/18 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:\therefore{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{7}\:{and}\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 161162 by metamorfose last updated on 13/Dec/21 $$\left.{f}\::\right]\mathrm{0},+\infty\left[\rightarrow\right]\mathrm{0},+\infty\left[\:{is}\:{convex}\:{function}\right. \\ $$$${for}\:{n}\geqslant\mathrm{2}\:{an}\:{integer}\:,\:{prove}\:: \\ $$$$\left({f}\left(\mathrm{1}\right)^{{f}\left(\mathrm{1}\right)} {f}\left(\mathrm{2}\right)^{{f}\left(\mathrm{2}\right)} …{f}\left({n}\right)^{{f}\left({n}\right)} \right)^{\frac{\mathrm{1}}{{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+…+{f}\left({n}\right)}} +\left({f}\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)…{f}\left({n}\right)\right)^{\frac{\mathrm{1}}{{n}}} \leqslant{f}\left(\mathrm{1}\right)+{f}\left({n}\right) \\ $$ Terms of Service Privacy…
Question Number 161159 by daus last updated on 13/Dec/21 Answered by Rasheed.Sindhi last updated on 13/Dec/21 $${x}=\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\:;{y}\neq\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\right)^{\mathrm{2}} −\mathrm{5}\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\right){y}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\…
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Question Number 161150 by mathlove last updated on 13/Dec/21 Commented by MJS_new last updated on 13/Dec/21 $${y}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$${y}=\sqrt{\mathrm{2}}\mid{x}\mid\:\Rightarrow\:{xy}=\sqrt{\mathrm{2}}\mid{x}\mid{x}=\begin{cases}{{x}<\mathrm{0};\:−\sqrt{\mathrm{2}}{x}^{\mathrm{2}} }\\{{x}\geqslant\mathrm{0};\:\sqrt{\mathrm{2}}{x}^{\mathrm{2}} }\end{cases} \\ $$ Terms…
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Question Number 95608 by bobhans last updated on 26/May/20 Commented by john santu last updated on 26/May/20 $$\mathrm{another}\:\mathrm{way} \\ $$$$\mathrm{set}\:{x}+{y}\:=\:{h}\:{and}\:{xy}\:=\:{g} \\ $$$$\Rightarrow{xy}\:\left({x}+{y}\right)\:=\:\mathrm{30}\:\&\:{g}\:+\:{h}\:=\:\mathrm{11} \\ $$$$\Rightarrow{gh}\:=\:\mathrm{30}\:;\:{g}\:=\:\mathrm{11}−{h} \\…
Question Number 161123 by HongKing last updated on 12/Dec/21 $$\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}}}…