Question Number 147893 by mathdanisur last updated on 24/Jul/21 $${x}\:+\:\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{4}\:\:\Rightarrow\:\:{x}\:+\:\mathrm{1}\:-\:\frac{\mathrm{1}}{{x}}\:=\:? \\ $$ Commented by aliibrahim1 last updated on 24/Jul/21 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\Rrightarrow\:\:\:\:\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{9}…
Question Number 147892 by mathdanisur last updated on 24/Jul/21 $${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 $$\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}<\frac{{x}^{\mathrm{2}} }{\mathrm{4}}…?\:\forall{x}>\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{4}\left({x}−\mathrm{1}\right)<{x}^{\mathrm{2}}…
Question Number 147894 by mathdanisur last updated on 24/Jul/21 $${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} \\ $$$$\Rightarrow{x}^{\mathrm{2}}…
Question Number 82356 by M±th+et£s last updated on 20/Feb/20 Commented by M±th+et£s last updated on 20/Feb/20 $${show}\:{that} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 147877 by Tawa11 last updated on 24/Jul/21 $$\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 $$\mathrm{Welcome}\:\mathrm{back}.\:\mathrm{Tawa}. \\ $$🙂🙂…
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 $${a}=\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:+\:\mathrm{10}}\:,{b}=\:\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{3}}\:−\:\mathrm{10}} \\…
Question Number 147878 by Tawa11 last updated on 24/Jul/21 $$\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) \\ $$$$\mathrm{S}\:=\:\mathrm{3}\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 16794 by tawa tawa last updated on 26/Jun/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17 $${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}} \\ $$$${a}+{c}={c}−\mathrm{2}+{c}=\mathrm{2}\left({c}−\mathrm{1}\right)=\mathrm{2}{b},{c}−{a}=\mathrm{2} \\…
Question Number 82307 by Power last updated on 20/Feb/20 Commented by Tony Lin last updated on 20/Feb/20 $$\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\…
Question Number 147842 by mathdanisur last updated on 23/Jul/21 $${x}\:;\:{y}\:;\:{z}\:>\:\mathrm{0} \\ $$$$\begin{cases}{{x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \:=\:\mathrm{3}}\\{{y}+{z}^{\mathrm{2}} +{x}^{\mathrm{3}} \:=\:\mathrm{3}}\\{{z}+{x}^{\mathrm{2}} +{y}^{\mathrm{3}} \:=\:\mathrm{3}}\end{cases}\:\:\Rightarrow\:\:{x}\:;\:{y}\:;\:{z}\:=\:? \\ $$ Commented by mr W last…