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Category: Algebra

if-x-y-z-gt-1-then-x-1-y-1-z-1-x-1-y-1-z-1-lt-xyz-8-

Question Number 147892 by mathdanisur last updated on 24/Jul/21 $${if}\:\:\:{x};{y};{z}>\mathrm{1}\:\:\:{then}: \\ $$$$\sqrt{\frac{\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\left({z}−\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)}\:}\:<\:\frac{{xyz}}{\mathrm{8}} \\ $$ Answered by mindispower last updated on 24/Jul/21 $$\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}<\frac{{x}^{\mathrm{2}} }{\mathrm{4}}…?\:\forall{x}>\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{4}\left({x}−\mathrm{1}\right)<{x}^{\mathrm{2}}…

x-2-3x-4-0-x-2-16-x-2-

Question Number 147894 by mathdanisur last updated on 24/Jul/21 $${x}^{\mathrm{2}} \:-\:\mathrm{3}{x}\:+\:\mathrm{4}\:=\:\mathrm{0}\:\:\Rightarrow\:\:{x}^{\mathrm{2}} \:+\:\frac{\mathrm{16}}{{x}^{\mathrm{2}} }\:=\:? \\ $$ Answered by mindispower last updated on 24/Jul/21 $$\Leftrightarrow{x}+\frac{\mathrm{4}}{{x}}=\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{2}}…

Find-the-sum-to-close-form-1-1-2-3-1-4-5-6-1-7-8-9-

Question Number 147877 by Tawa11 last updated on 24/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{close}\:\mathrm{form}. \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:\:\:+\:\:\:…\:\:\: \\ $$ Commented by Tinku Tara last updated on 24/Jul/21 $$\mathrm{Welcome}\:\mathrm{back}.\:\mathrm{Tawa}. \\ $$🙂🙂…

if-6-3-10-1-3-6-3-10-1-3-x-find-x-3-10-3x-

Question Number 147882 by mathdanisur last updated on 24/Jul/21 $${if}\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:+\:\mathrm{10}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:−\:\mathrm{10}}\:=\:\boldsymbol{{x}} \\ $$$${find}\:\:\:\frac{{x}^{\mathrm{3}} }{\mathrm{10}−\mathrm{3}{x}}\:=\:? \\ $$ Answered by mr W last updated on 24/Jul/21 $${a}=\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:+\:\mathrm{10}}\:,{b}=\:\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:−\:\mathrm{10}} \\…

Find-the-sum-to-n-term-1-2-3-4-5-6-7-8-9-

Question Number 147878 by Tawa11 last updated on 24/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\:\mathrm{n}\:\:\mathrm{term}:\:\:\:\:\:\mathrm{1}.\mathrm{2}.\mathrm{3}\:\:+\:\:\mathrm{4}.\mathrm{5}.\mathrm{6}\:\:+\:\:\mathrm{7}.\mathrm{8}.\mathrm{9}\:\:+\:\:… \\ $$ Answered by Olaf_Thorendsen last updated on 24/Jul/21 $$\mathrm{S}\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{k}−\mathrm{2}\right)\left(\mathrm{3}{k}−\mathrm{1}\right)\left(\mathrm{3}{k}\right) \\ $$$$\mathrm{S}\:=\:\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}}…

Question-16794

Question Number 16794 by tawa tawa last updated on 26/Jun/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17 $${c}−{x}={t}^{\mathrm{3}} ,{x}−{a}={s}^{\mathrm{3}} \Rightarrow{x}=\frac{{s}^{\mathrm{3}} −{t}^{\mathrm{3}} }{\mathrm{2}}+\frac{{a}+{c}}{\mathrm{2}} \\ $$$${a}+{c}={c}−\mathrm{2}+{c}=\mathrm{2}\left({c}−\mathrm{1}\right)=\mathrm{2}{b},{c}−{a}=\mathrm{2} \\…

Question-82307

Question Number 82307 by Power last updated on 20/Feb/20 Commented by Tony Lin last updated on 20/Feb/20 $$\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\…

prove-that-cos-2-log-cos-sin-cos-sin-1-2-sin-2-log-tan-pi-4-1-2-log-cos-2-

Question Number 147835 by peter frank last updated on 23/Jul/21 $${prove}\:{that}\: \\ $$$$\:\:\int\mathrm{cos}\:\mathrm{2}\theta{log}\left(\frac{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta{log}\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\left(\mathrm{cos}\:\mathrm{2}\theta\right)\right. \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com