Question Number 175154 by Shrinava last updated on 21/Aug/22 Commented by mr W last updated on 21/Aug/22 $${i}\:{don}'{t}\:{think}\:{there}\:{are}\:{unique}\:{solutions}. \\ $$ Answered by MJS_new last updated…
Question Number 175148 by Shrinava last updated on 20/Aug/22 Commented by mr W last updated on 21/Aug/22 $${wrong}! \\ $$$${for}\:{a}=−\mathrm{1},\:{b}=\mathrm{1} \\ $$$$\mathrm{2}^{{b}} −\mathrm{2}^{{a}} =\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1}.\mathrm{5}>\frac{\mathrm{9}\:\mathrm{log}\:\mathrm{2}}{\mathrm{4}}\approx\mathrm{0}.\mathrm{677} \\…
Question Number 109611 by mnjuly1970 last updated on 24/Aug/20 Commented by Rasheed.Sindhi last updated on 25/Aug/20 $$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{3}} =\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} +\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} }+\mathrm{3}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 44062 by peter frank last updated on 20/Sep/18 Answered by MJS last updated on 21/Sep/18 $$\mathrm{roots}\:{r}_{\mathrm{1}} ,\:{r}_{\mathrm{2}} \:\Rightarrow\:\left({x}−{r}_{\mathrm{1}} \right)\left({x}−{r}_{\mathrm{2}} \right)=\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{2}} −\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right){x}+{r}_{\mathrm{1}}…
Question Number 175132 by mnjuly1970 last updated on 20/Aug/22 $$ \\ $$$$\:\:\:\:\:\:\prec\:\:\boldsymbol{{Question}}−\:\boldsymbol{{algebra}}\:\succ \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{Count}}\:\boldsymbol{{the}}\:\boldsymbol{{number}}\:\boldsymbol{{of}}\:\:\:''\:\boldsymbol{{zero}}\:\boldsymbol{{divisor}}\:'' \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{of}}\:\:\boldsymbol{{ring}}\:,\:\:\:\left(\:\mathbb{Z}_{\:\mathrm{45}} \:,\: \:,\:\oplus\:\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\:\boldsymbol{{m}}.\boldsymbol{{n}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−−−−\:\:\:\: \\ $$ Commented…
Question Number 109595 by nimnim last updated on 24/Aug/20 $${For}\:{any}\:{Real}\:{numbers}\:{x},{y}\:{and}\:{z}, \\ $$$$\:{if}\:\:\left({x}+{y}+{z}\right)=\mathrm{2},\:{then}\:{prove}\:{that} \\ $$$$\:\:\:\:\:\:\:\:{xyz}\geqslant\mathrm{8}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right) \\ $$ Commented by 1549442205PVT last updated on 25/Aug/20 $$\mathrm{This}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{only}\:\mathrm{true}\:\mathrm{for} \\…
Question Number 175121 by Michaelfaraday last updated on 20/Aug/22 Answered by BaliramKumar last updated on 20/Aug/22 $$\left(\mathrm{3}^{−\mathrm{1}} \right)^{\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{16}−\mathrm{2}{x}^{\mathrm{3}} }} \:=\:\left(\mathrm{3}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}{x}}} \\ $$$$\mathrm{3}^{\left(\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{2}{x}^{\mathrm{3}}…
Question Number 109577 by bemath last updated on 24/Aug/20 $$\:\:\:\mathrm{G}{iven}\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} =\mathrm{133} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{and}\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$\:\:{then}\:{what}\:{is}\:{the}\:{value}\:{of}\:{xy}\:? \\ $$ Commented by bemath…
Question Number 44038 by Tawa1 last updated on 20/Sep/18 $$\mathrm{How}\:\mathrm{many}\:\mathrm{times}\:\mathrm{does}\:\mathrm{the}\:\mathrm{digit}\:\mathrm{6}\:\mathrm{appear}\:\mathrm{when}\:\mathrm{writing}\:\mathrm{from}\:\:\mathrm{6}\:\mathrm{to}\:\mathrm{400}\:? \\ $$ Answered by $@ty@m last updated on 20/Sep/18 $$\mathrm{6},\:\mathrm{16},\mathrm{26},\mathrm{36},\mathrm{46},\mathrm{56}, \\ $$$$\mathrm{60},\mathrm{61},\mathrm{62},\mathrm{63},\mathrm{64},\mathrm{65},\mathrm{66},\mathrm{67},\mathrm{68},\mathrm{69}, \\ $$$$\mathrm{76},\mathrm{86},\mathrm{96} \\…
Question Number 109569 by bemath last updated on 24/Aug/20 Answered by john santu last updated on 24/Aug/20 $$\:\:\:\frac{\leftrightharpoons{JS}\rightleftharpoons}{\spadesuit} \\ $$$${let}\:\:\frac{{x}−\mathrm{1}}{{x}}={t}\:\rightarrow{x}=\frac{\mathrm{1}}{\mathrm{1}−{t}} \\ $$$$\left(\mathrm{1}\right){f}\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)+{f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{t}}+\mathrm{1}=\frac{\mathrm{2}−{t}}{\mathrm{1}−{t}} \\ $$$${set}\:{x}\:=\:\frac{{t}−\mathrm{1}}{{t}} \\…