Question Number 145641 by physicstutes last updated on 06/Jul/21 $$\mathrm{Let}\:\mathrm{g}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{given}\:\mathrm{by}\:\mathrm{g}\left({x}\right)\:=\:\mathrm{3}\:+\:\mathrm{4}{x}\:.\mathrm{Prove}\:\mathrm{by}\:\mathrm{induction} \\ $$$$\mathrm{that},\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:{n},\: \\ $$$$\mathrm{g}^{{n}} \left({x}\right)\:=\:\left(\mathrm{4}^{{n}} −\mathrm{1}\right)\:+\:\mathrm{4}^{{n}} \left({x}\right). \\ $$$$\mathrm{If}\:\mathrm{for}\:\mathrm{every}\:\mathrm{positive}\:\mathrm{integer}\:{k},\:\mathrm{we}\:\mathrm{inteprete}\:\mathrm{g}^{−{k}} \:\mathrm{as}\:\mathrm{the}\:\mathrm{inverse} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{g}^{{k}} .\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{above}\:\mathrm{formula}\:\mathrm{holds}\:\mathrm{alsl} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{negative}\:\mathrm{integers}\:{n}.…
Question Number 80093 by TawaTawa last updated on 30/Jan/20 $$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\:\:\:\:\:\mathrm{x}^{\sqrt{\mathrm{y}}} \:\:\:=\:\:\mathrm{64} \\ $$$$\:\:\:\:\:\mathrm{y}^{\sqrt{\mathrm{x}}} \:\:\:=\:\mathrm{81} \\ $$ Answered by MJS last updated on 30/Jan/20…
Question Number 14559 by tawa tawa last updated on 02/Jun/17 $$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x} \\ $$$$\frac{\mathrm{6x}\:+\:\mathrm{2a}\:+\:\mathrm{3b}\:+\:\mathrm{c}\:}{\mathrm{6x}\:+\:\mathrm{2a}\:−\:\mathrm{3b}\:−\:\mathrm{c}}\:=\:\frac{\mathrm{2x}\:+\:\mathrm{6a}\:+\:\mathrm{b}\:+\:\mathrm{3c}}{\mathrm{2x}\:+\:\mathrm{6a}\:−\:\mathrm{b}\:−\:\mathrm{3c}} \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/Jun/17 $$\frac{\left(\mathrm{6}{x}+\mathrm{2}{a}+\mathrm{3}{b}+{c}\right)+\left(\mathrm{6}{x}+\mathrm{2}{a}−\mathrm{3}{b}−{c}\right)}{\left(\mathrm{6}{x}+\mathrm{2}{a}+\mathrm{3}{b}+{c}\right)−\left(\mathrm{6}{x}+\mathrm{2}{a}−\mathrm{3}{b}−{c}\right)}= \\ $$$$=\frac{\left(\mathrm{2}{x}+\mathrm{6}{a}+{b}+\mathrm{3}{c}\right)+\left(\mathrm{2}{x}+\mathrm{6}{a}−{b}−\mathrm{3}{c}\right)}{\left(\mathrm{2}{x}+\mathrm{6}{a}+{b}+\mathrm{3}{c}\right)−\left(\mathrm{2}{x}+\mathrm{6}{a}−{b}−\mathrm{3}{c}\right)}\Rightarrow…
Question Number 145626 by mathdanisur last updated on 06/Jul/21 Commented by mr W last updated on 06/Jul/21 $$\mid{a}+{b}\mid\leqslant\mid{a}\mid+\mid{b}\mid \\ $$$$\frac{\mid{a}\mid+\mid{b}\mid}{\mid{a}+{b}\mid}\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mid{a}\mid+\mid{b}\mid}{\mid{a}+{b}\mid}\geqslant\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}=\frac{\mathrm{3}}{\mathrm{2}}={min} \\ $$ Commented…
Question Number 80084 by behi83417@gmail.com last updated on 30/Jan/20 $$\:\:−\mathrm{1}=\left(−\mathrm{1}\right)^{\mathrm{1}} =\left(−\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{2}}} =\left(\left(−\mathrm{1}\right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} = \\ $$$$=\sqrt{\mathrm{1}}=\mathrm{1}\:\: \\ $$$$\mathrm{what}\:\mathrm{do}\:\mathrm{you}\:\mathrm{think}\:\mathrm{about}\:\mathrm{this}? \\ $$ Commented by key of…
Question Number 145615 by loveineq last updated on 06/Jul/21 $$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:{abc}\:=\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} }{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{b}^{\mathrm{3}} }{\left({b}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{c}^{\mathrm{3}} }{\left({c}+\mathrm{1}\right)^{\mathrm{2}} }\:\geqslant\frac{{a}+{b}+{c}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Commented by justtry last updated…
Question Number 80068 by Pratah last updated on 30/Jan/20 Commented by mr W last updated on 30/Jan/20 $${sir}\:{Pratah}: \\ $$$${are}\:{you}\:{student}?\:{teacher}?\:{professor}? \\ $$$${what}'{s}\:{the}\:{source}\:{of}\:{your}\:{questions}? \\ $$ Commented…
Question Number 14535 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17 $${hello}\:{dears}. \\ $$$${this}\:{is}\:{an}\:{ancient}\:{system}\:{of}\:{equiations} \\ $$$${that}\:{have}\:{not}\:{solved}\:{by}\:{any}\:{one}\:{yet}. \\ $$$${but}\:{i}\:{think}\:{by}\:{studing}\:{several}\:{ways} \\ $$$${that}\:{used}\:{for}\:{solveing}\:{Q}.\mathrm{14157},{now}\:…
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Question Number 80053 by mr W last updated on 30/Jan/20 $${Find}\:{integer}\:{x},\:{y}\:{such}\:{that} \\ $$$$\mathrm{2}^{{x}} −{y}^{\mathrm{2}} =\mathrm{615} \\ $$ Commented by john santu last updated on 30/Jan/20…