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Question Number 195118 by York12 last updated on 25/Jul/23 $${a},{b},{c}>\mathrm{0}\:\&\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{3} \\ $$$${prove}\:{that} \\ $$$$\frac{{a}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{b}}{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{c}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqslant\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{a}+{b}+{c}}{{ab}+{bc}+{ac}}\right)^{\mathrm{2}} \\ $$ Commented by York12…

Question Number 195114 by mustafazaheen last updated on 24/Jul/23 $$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\underset{\mathrm{k}=\mathrm{3}} {\overset{\mathrm{55}} {\sum}}\mathrm{k}=? \\ $$$$\left.\mathrm{1}\left.\right)\left.\mathrm{5}\left.\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\right)\mathrm{9k}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{53k}\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\right)\mathrm{45} \\ $$$$ \\ $$$$ \\ $$ Answered…

Question Number 195083 by sciencestudentW last updated on 23/Jul/23 $${x}=\sqrt{\mathrm{5}}−\mathrm{2}\:\:\:\:\: \\ $$$${x}+\frac{\mathrm{1}}{{x}}=? \\ $$ Answered by Rasheed.Sindhi last updated on 23/Jul/23 $${x}=\sqrt{\mathrm{5}}−\mathrm{2}\:\:,\:\:{x}+\frac{\mathrm{1}}{{x}}=? \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}−\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}−\mathrm{2}}\centerdot\frac{\sqrt{\mathrm{5}}\:+\mathrm{2}}{\:\sqrt{\mathrm{5}}\:+\mathrm{2}} \\…

Question Number 195038 by sciencestudentW last updated on 22/Jul/23 $${prove}\:{that}\:{x}+{y}=\frac{\mathrm{1}}{{x}−{y}} \\ $$ Commented by Frix last updated on 22/Jul/23 $$\mathrm{It}'\mathrm{s}\:\mathrm{wrong},\:\mathrm{i}.\mathrm{e}.\:{x}=\mathrm{2}\wedge{y}=\mathrm{3}\:\Rightarrow\: \\ $$$${x}+{y}=\mathrm{5}\:\mathrm{but}\:\frac{\mathrm{1}}{{x}−{y}}=−\mathrm{1} \\ $$$$ \\…

Question Number 195020 by tri26112004 last updated on 22/Jul/23 $${y}={x}^{{x}^{{x}…} } \\ $$$$=>\:\frac{{dy}}{{dx}}=¿ \\ $$ Answered by Frix last updated on 22/Jul/23 $${y}={x}^{{x}^{{x}…} } \\…