Question Number 191439 by Shrinava last updated on 24/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125900 by Mathgreat last updated on 15/Dec/20 Commented by Mathgreat last updated on 15/Dec/20 $$?????? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 191430 by TechnicalDev last updated on 24/Apr/23 $${factorise}\:\mathrm{4}{x}^{\mathrm{4}} +\mathrm{81} \\ $$ Answered by som(math1967) last updated on 24/Apr/23 $$\mathrm{4}{x}^{\mathrm{4}} +\mathrm{81} \\ $$$$=\left(\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}}…
Question Number 125890 by geraldinebalasta last updated on 15/Dec/20 $$\left(\mathbb{Z}^{+,\ast} \right)\:{where}\:{x}^{\ast} {y}=\:{x}−{y}\:{for}\:{all}\:{x},{y}\:\in\:{z}^{+} . \\ $$ Commented by talminator2856791 last updated on 15/Dec/20 $$\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}? \\ $$…
Question Number 125873 by Mathgreat last updated on 14/Dec/20 Answered by MJS_new last updated on 15/Dec/20 $${t}=\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:=\mathrm{2sin}\:{x}\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:=\begin{cases}{\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}};\:\frac{\sqrt{\mathrm{1}+{t}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{1}−{t}}}{\mathrm{2}}}\\{\frac{\pi}{\mathrm{4}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}};\:\frac{\sqrt{\mathrm{1}+{t}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{1}−{t}}}{\mathrm{2}}}\end{cases} \\ $$$$\rightarrow\:{dx}=\begin{cases}{\frac{{dt}}{\mathrm{2}\sqrt{\mathrm{1}+{t}}\sqrt{\mathrm{1}−{t}}}}\\{−\frac{{dt}}{\mathrm{2}\sqrt{\mathrm{1}+{t}}\sqrt{\mathrm{1}−{t}}}}\end{cases} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}}…
Question Number 125868 by I want to learn more last updated on 14/Dec/20 Answered by mindispower last updated on 14/Dec/20 $$\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{y}}\geqslant\frac{\mathrm{2}}{{x}+{y}}.\Leftrightarrow\left({x}+{y}\right)^{\mathrm{2}} \geqslant\mathrm{4}{xy},{x},{y}>\mathrm{0} \\ $$$$\Leftrightarrow\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\:{true}…
Question Number 60330 by ANTARES VY last updated on 20/May/19 Commented by ANTARES VY last updated on 20/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125864 by MathSh last updated on 14/Dec/20 $$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{4}} }+\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{3}}{\mathrm{1}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{4}} }+…+\frac{\mathrm{50}}{\mathrm{1}+\mathrm{50}^{\mathrm{2}} +\mathrm{50}^{\mathrm{4}} }=? \\ $$ Answered by Dwaipayan Shikari last…
Question Number 191393 by Mingma last updated on 23/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191395 by Mingma last updated on 23/Apr/23 Answered by Rasheed.Sindhi last updated on 23/Apr/23 $${xy}+{yz}+{zx}=\mathrm{0};\:\:\frac{{y}+{z}}{{x}}+\frac{{x}+{z}}{{y}}+\frac{{x}+{y}}{{z}}=? \\ $$$$\frac{{y}+{z}}{{x}}+\frac{{x}+{z}}{{y}}+\frac{{x}+{y}}{{z}} \\ $$$$=\frac{{y}+{z}}{{x}}+\mathrm{1}+\frac{{x}+{z}}{{y}}+\mathrm{1}+\frac{{x}+{y}}{{z}}+\mathrm{1}−\mathrm{3} \\ $$$$=\frac{{x}+{y}+{z}}{{x}}+\frac{{x}+{y}+{z}}{{y}}+\frac{{x}+{y}+{z}}{{z}}−\mathrm{3} \\ $$$$=\left({x}+{y}+{z}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)−\mathrm{3}…