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 $${f}\left({f}\left({x}\right)\right)=\frac{\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}−\mathrm{3}}{\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}+\mathrm{1}}\:=\:\frac{{x}−\mathrm{3}−\mathrm{3}{x}−\mathrm{3}}{{x}−\mathrm{3}+{x}+\mathrm{1}} \\ $$$${f}\left({f}\left({x}\right)\right)=\:\frac{−\mathrm{2}{x}−\mathrm{6}}{\mathrm{2}{x}−\mathrm{2}}\:=\:\frac{−{x}−\mathrm{3}}{{x}−\mathrm{1}}={f}^{−\mathrm{1}} \left({x}\right)…
Question Number 142668 by mathdanisur last updated on 03/Jun/21 $$\underset{{n}\rightarrow\infty} {{lim}}\left(\frac{{n}+\mathrm{6}}{{n}}\right)^{\frac{\mathrm{6}}{{n}}} =\:? \\ $$ Commented by iloveisrael last updated on 04/Jun/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{n}+\mathrm{6}}{{n}}\right)^{\frac{\mathrm{6}}{{n}}} \:= \\…
Question Number 77119 by necxxx last updated on 03/Jan/20 $${suppose}\:{the}\:{equations}\:{x}^{\mathrm{2}} +{px}+\mathrm{4}=\mathrm{0} \\ $$$${and}\:{x}^{\mathrm{2}} +{qx}+\mathrm{3}=\mathrm{0}\:\:{have}\:{a}\:{common}\:{root}, \\ $$$${write}\:{this}\:{root}\:{in}\:{terms}\:{of}\:{the}\:{other}\:{root}. \\ $$ Answered by jagoll last updated on 03/Jan/20…
Question Number 142647 by iloveisrael last updated on 03/Jun/21 $$\:\frac{\mathrm{1}}{\mathrm{2018}}−\frac{\mathrm{2}}{\mathrm{2018}}+\frac{\mathrm{3}}{\mathrm{2018}}−\frac{\mathrm{4}}{\mathrm{2018}}+…−\frac{\mathrm{2016}}{\mathrm{2018}}+\frac{\mathrm{2017}}{\mathrm{2018}}=? \\ $$ Answered by MJS_new last updated on 03/Jun/21 $$\underset{{j}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\left(\mathrm{2}{j}+\mathrm{1}\right)−\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{j}\right)={n}+\mathrm{1} \\…