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Category: Algebra

log-2-3-x-log-3-5-y-lg6-

Question Number 144120 by mathdanisur last updated on 21/Jun/21 $${log}_{\mathrm{2}} \mathrm{3}\:=\:{x}\:,\:{log}_{\mathrm{3}} \mathrm{5}\:=\:{y}\:,\:{lg}\mathrm{6}\:=\:? \\ $$ Answered by Ar Brandon last updated on 21/Jun/21 $$\mathrm{log}_{\mathrm{3}} \mathrm{5}=\mathrm{y}\Rightarrow\mathrm{log}_{\mathrm{2}} \mathrm{5}=\mathrm{xy}…

Let-a-b-c-gt-0-and-a-b-c-3-Prove-that-ab-1-ab-c-bc-1-bc-a-ca-1-ca-b-a-b-c-

Question Number 144118 by loveineq last updated on 21/Jun/21 $$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{{ab}}+\mathrm{1}}{\:\sqrt{{ab}}+\sqrt{{c}}}+\frac{\sqrt{{bc}}+\mathrm{1}}{\:\sqrt{{bc}}+\sqrt{{a}}}+\frac{\sqrt{{ca}}+\mathrm{1}}{\:\sqrt{{ca}}+\sqrt{{b}}}\:\geqslant\:\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

MrW1-Before-we-concluded-that-x-0-m-y-0-n-1-sgn-x-x-If-you-do-x-0-1-y-0-1-1-sgn-x-x-x-0-1-y-0-1-1-sgn-x-LCM-x-y-y-1-sgn-0-LCM-0-0-0-1-s

Question Number 13034 by FilupS last updated on 12/May/17 $$\mathrm{MrW1} \\ $$$$\: \\ $$$$\mathrm{Before}\:\mathrm{we}\:\mathrm{concluded}\:\mathrm{that}: \\ $$$$\Phi=\underset{{x}=\mathrm{0}} {\overset{{m}} {\sum}}\underset{{y}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\mathrm{sgn}\left({x}−{x}'\right)\right) \\ $$$$\: \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{do}: \\…

Find-the-roots-of-the-equation-bx-3-3b-2-x-2-2-5b-3-x-20-0-

Question Number 78568 by TawaTawa last updated on 18/Jan/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{bx}^{\mathrm{3}} \:−\:\left(\mathrm{3b}\:+\:\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{5b}\:−\:\mathrm{3}\right)\mathrm{x}\:+\:\mathrm{20}\:\:=\:\:\mathrm{0} \\ $$ Answered by key of knowledge last updated on 18/Jan/20…

if-x-gt-0-r-pq-1-p-q-then-1-rx-1-qx-p-1-px-q-1-x-r-

Question Number 144095 by mathdanisur last updated on 21/Jun/21 $${if}\:\:{x}>\mathrm{0}\:,\:{r}={pq}\:,\:\mathrm{1}\leqslant{p}\leqslant{q}\:\:{then}: \\ $$$$\mathrm{1}+{rx}\:\leqslant\:\left(\mathrm{1}+{qx}\right)^{\boldsymbol{{p}}} \:\leqslant\:\left(\mathrm{1}+{px}\right)^{\boldsymbol{{q}}} \:\leqslant\:\left(\mathrm{1}+{x}\right)^{\boldsymbol{{r}}} \\ $$ Answered by mitica last updated on 21/Jun/21 $$\left(\mathrm{1}+{x}\right)^{{r}} =\left[\left(\mathrm{1}+{x}\right)^{{p}}…