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Author: Tinku Tara

Show-that-x-x-1-3x-1-gt-1-given-that-x-gt-1-3-

Question Number 984 by 112358 last updated on 13/May/15 $${Show}\:{that}\:\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{3}{x}−\mathrm{1}}>\mathrm{1}\:{given}\:{that}\:{x}>\frac{\mathrm{1}}{\mathrm{3}}\:. \\ $$ Commented by prakash jain last updated on 13/May/15 $$\mathrm{For}\:{x}=\mathrm{1}\:\mathrm{LHS}=\frac{\mathrm{1}×\mathrm{2}}{\mathrm{2}}=\mathrm{1}\ngtr\mathrm{1} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{should}\:\mathrm{be} \\ $$$$\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{3}{x}−\mathrm{1}}\geqslant\mathrm{1}…

Question-66518

Question Number 66518 by Masumsiddiqui399@gmail.com last updated on 16/Aug/19 Commented by mathmax by abdo last updated on 16/Aug/19 $${let}\:{f}\left({x}\right)=\frac{{x}\sqrt{{x}}−{a}\sqrt{{a}}}{{x}−{a}}\:\:\:{cha}\mathrm{7}{gement}\:{x}−{a}={t}\:{give} \\ $$$${lim}_{{x}\rightarrow{a}} {f}\left({x}\right)\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{\left({t}+{a}\right)\sqrt{{t}+{a}}−{a}\sqrt{{a}}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}}…

tan-x-tan-y-z-a-tan-y-tan-z-x-b-tan-z-tan-x-y-c-for-wich-values-of-a-b-c-the-system-have-solutions-x-y-z-a-b-c-R-6-

Question Number 976 by 123456 last updated on 11/May/15 $$\begin{cases}{\mathrm{tan}\:{x}\:\mathrm{tan}\:\left({y}−{z}\right)={a}}\\{\mathrm{tan}\:{y}\:\mathrm{tan}\:\left({z}−{x}\right)={b}}\\{\mathrm{tan}\:{z}\:\mathrm{tan}\:\left({x}−{y}\right)={c}}\end{cases} \\ $$$$\mathrm{for}\:\mathrm{wich}\:\mathrm{values}\:\mathrm{of}\:{a},{b},{c}\:\mathrm{the}\:\mathrm{system} \\ $$$$\mathrm{have}\:\mathrm{solutions}? \\ $$$$\left({x},{y},{z},{a},{b},{c}\right)\in\mathbb{R}^{\mathrm{6}} \\ $$ Commented by 123456 last updated on 11/May/15…

Question-66508

Question Number 66508 by Masumsiddiqui399@gmail.com last updated on 16/Aug/19 Commented by Prithwish sen last updated on 16/Aug/19 $$\left(\sqrt{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}\right)^{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}} =\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}+^{\mathrm{3}} \sqrt{\mathrm{2x}+\mathrm{4}}} \\ $$$$\because\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\:=\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\therefore\:\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}=\mathrm{x}+^{\mathrm{3}} \sqrt{\mathrm{2x}+\mathrm{1}}\:\:\Rightarrow\mathrm{x}=\mathrm{2}…