Question Number 142740 by mathdanisur last updated on 04/Jun/21 $$\frac{{a}}{{b}+{c}}\:=\:\frac{{b}}{{a}+{c}}\:=\:\frac{{c}}{{a}+{b}}\:\:;\:\:{a}\centerdot{b}\centerdot{c}\:=\:\mathrm{12} \\ $$$$\left({b}+{c}\right)\centerdot\left({a}+{c}\right)\centerdot\left({a}+{b}\right)\:=\:? \\ $$ Commented by MJS_new last updated on 05/Jun/21 $$\mathrm{I}\:\mathrm{found}\:\mathrm{2}\:\mathrm{possibilities} \\ $$$$\left(\mathrm{1}\right)\:{a}={b}={c}=\sqrt[{\mathrm{3}}]{\mathrm{12}}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{96} \\…
Question Number 11672 by @ANTARES_VY last updated on 29/Mar/17 $$\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{6}\boldsymbol{\mathrm{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{Solves}}… \\ $$ Commented by mrW1 last updated on 29/Mar/17 $${see}\:{general}\:{solution}: \\ $$…
Question Number 11661 by @ANTARES_VY last updated on 29/Mar/17 $$\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{6}\boldsymbol{\mathrm{x}}+\mathrm{1}=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{Solves}}…\:\:\boldsymbol{\mathrm{equation}} \\ $$$$\boldsymbol{\mathrm{x}}_{\mathrm{1}} =?\:\:\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} =? \\ $$ Answered by Joel576 last updated on…
Question Number 142725 by mathlove last updated on 04/Jun/21 Commented by Olaf_Thorendsen last updated on 04/Jun/21 $${https}://{peterjamesthomas}.{com}/{category}/{general}/{social}−{media}/?{hcb}=\mathrm{1} \\ $$ Commented by MJS_new last updated on…
Question Number 11648 by uni last updated on 29/Mar/17 $$\mid\mathrm{x}\mid<\mathrm{l} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{x}^{\mathrm{n}} =? \\ $$ Answered by Nayon last updated on 29/Mar/17 $$…
Question Number 77183 by jagoll last updated on 04/Jan/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{x}\: \\ $$$$\mathrm{satisfy}\:\mathrm{inequality}\: \\ $$$$\mathrm{3}^{\mathrm{x}^{\mathrm{2}} } ×\:\mathrm{5}^{\mathrm{x}−\mathrm{1}} \:\geqslant\:\mathrm{3} \\ $$ Answered by john santu last updated…
Question Number 142710 by loveineq last updated on 04/Jun/21 $$\mathrm{Let}\:{k}\:\mathrm{be}\:\mathrm{non}-\mathrm{negative}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and}\:{n}\:\in\:\mathrm{N}^{+} \geqslant\mathrm{1}.\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{4}−{k}}{\mathrm{4}+{k}}\right)\left(\frac{\mathrm{12}−{k}}{\mathrm{12}+{k}}\right)…\left[\frac{\mathrm{2}{n}\left({n}+\mathrm{1}\right)−{k}}{\mathrm{2}{n}\left({n}+\mathrm{1}\right)+{k}}\right]\:\leqslant\:\frac{{n}+{k}+\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({k}+\mathrm{1}\right)} \\ $$$$ \\ $$ Answered by loveineq last updated on…
Question Number 11628 by uni last updated on 29/Mar/17 $$\mathrm{p}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{p}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{4x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{10} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=? \\ $$ Answered by sandy_suhendra last updated on 29/Mar/17 $$\mathrm{let}\:\mathrm{p}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\…
Question Number 77154 by Boyka last updated on 03/Jan/20 Commented by Boyka last updated on 04/Jan/20 $$\mathrm{sir}\:\:\:\mathrm{w}\:\:,\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power} \\ $$ Commented by mr W last updated…
Question Number 142674 by Snail last updated on 03/Jun/21 $${If}\:{f}\left({x}\right)=\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}\:{and}\:{h}\left({x}\right)={f}\left({f}\left({f}\left({f}\left(..\mathrm{2022}\:{times}\left({f}\left({x}\right)\right)\right)\right)\right)\right) \\ $$$${then}\:\:\:{h}\left(\mathrm{1}\right)\:=? \\ $$ Answered by iloveisrael last updated on 04/Jun/21 $${f}\left({f}\left({x}\right)\right)=\frac{\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}−\mathrm{3}}{\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}+\mathrm{1}}\:=\:\frac{{x}−\mathrm{3}−\mathrm{3}{x}−\mathrm{3}}{{x}−\mathrm{3}+{x}+\mathrm{1}} \\ $$$${f}\left({f}\left({x}\right)\right)=\:\frac{−\mathrm{2}{x}−\mathrm{6}}{\mathrm{2}{x}−\mathrm{2}}\:=\:\frac{−{x}−\mathrm{3}}{{x}−\mathrm{1}}={f}^{−\mathrm{1}} \left({x}\right)…