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Category: Algebra

1-2-a-b-a-b-min-

Question Number 145530 by mathdanisur last updated on 05/Jul/21 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mid{a}\mid\:+\:\mid{b}\mid}{\mid{a}\:+\:{b}\mid}\right)_{\boldsymbol{{min}}} =\:? \\ $$ Answered by puissant last updated on 07/Jul/21 $$\mid\mathrm{a}+\mathrm{b}\mid\leqslant\mid\mathrm{a}\mid+\mid\mathrm{b}\mid\:\Rightarrow\:\frac{\mid\mathrm{a}\mid+\mid\mathrm{b}\mid}{\mid\mathrm{a}+\mathrm{b}\mid}\:\geqslant\mathrm{1} \\ $$$$\Rightarrow\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mid\mathrm{a}\mid+\mid\mathrm{b}\mid}{\mid\mathrm{a}+\mathrm{b}\mid}\right)\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{take}\:\mathrm{min}=\frac{\mathrm{3}}{\mathrm{2}}..…

montrer-que-l-ensemble-des-suites-reelle-qui-verifie-la-relation-n-N-aU-n-2-bU-n-1-cU-n-0-1-est-un-espace-vectoriel-de-dimension-2-et-determiner-une-base-

Question Number 145522 by ArielVyny last updated on 05/Jul/21 $${montrer}\:{que}\:{l}'{ensemble}\:{des}\:{suites}\:{reelle}\:{qui} \\ $$$${verifie}\:{la}\:{relation}\:\forall{n}\in\mathbb{N} \\ $$$${aU}_{{n}+\mathrm{2}} +{bU}_{{n}+\mathrm{1}} +{cU}_{{n}} =\mathrm{0}\:\left(\mathrm{1}\right)\:\:{est}\:{un}\:{espace} \\ $$$${vectoriel}\:{de}\:{dimension}\:\mathrm{2} \\ $$$${et}\:{determiner}\:{une}\:{base}\: \\ $$$$ \\ $$…

Question-79974

Question Number 79974 by TawaTawa last updated on 29/Jan/20 Answered by Rio Michael last updated on 30/Jan/20 $$\:\boldsymbol{\mathrm{solution}} \\ $$$$\:\:\left[\left({p}\:\vee\:{q}\right)\:\wedge\:\left(\sim{p}\:\vee{r}\right)\right]\:\Rightarrow\:\left({q}\:\vee{r}\right) \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{from}\:\mathrm{known}\:\mathrm{facts}\:\mathrm{that}\: \\ $$$$\:−\left({p}\:\vee{q}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{contingency}\:\left(\mathrm{neither}\:\mathrm{a}\:\mathrm{tautology}\:\mathrm{nor}\:\mathrm{contradiction}\right) \\…

1-2-1-4-1-8-1-16-1-256-z-1-256-find-z-

Question Number 145499 by mathdanisur last updated on 05/Jul/21 $$\frac{\mathrm{1}}{\mathrm{2}}\:-\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\:-\:\frac{\mathrm{1}}{\mathrm{16}}\:+\:…\:-\:\frac{\mathrm{1}}{\mathrm{256}}\:=\:\frac{\boldsymbol{{z}}+\mathrm{1}}{\mathrm{256}} \\ $$$${find}\:\:\boldsymbol{{z}}=? \\ $$ Answered by gsk2684 last updated on 05/Jul/21 $$\mathrm{sum}\:\mathrm{of}\:\mathrm{8}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{G}.\mathrm{P}.\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{8}} }{\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}\right)=\frac{\mathrm{84}+\mathrm{1}}{\mathrm{256}}\:…

Solve-7-2-3y-x-y-x-4-y-equation-i-x-2-y-2-x-1-4-2xy-x-1-equation-ii-

Question Number 14398 by tawa tawa last updated on 31/May/17 $$\mathrm{Solve}:\: \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\:+\:\frac{\mathrm{3y}}{\mathrm{x}\:+\:\mathrm{y}}\:=\:\sqrt{\mathrm{x}}\:+\:\mathrm{4}\sqrt{\mathrm{y}}\:\:\:\:\:\:\:\:\:\:\:\:……….\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{4}\:+\:\mathrm{2xy}\left(\mathrm{x}\:−\:\mathrm{1}\right)\:\:\:\:……….\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on…