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Question Number 30221 by naka3546 last updated on 18/Feb/18 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 30206 by .none. last updated on 18/Feb/18 $$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}}} \\ $$$${x}=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$ Commented by abdo imad last updated on 18/Feb/18 $${let}\:{put}\:{a}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\:\Rightarrow{N}=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{a}}}=\frac{\mathrm{1}}{\frac{{a}−\mathrm{1}}{{a}}}\:=\frac{{a}}{{a}−\mathrm{1}}\:{but} \\ $$$${a}=\mathrm{1}−\frac{\mathrm{1}}{\frac{{x}}{{x}−\mathrm{1}}}=\mathrm{1}−\frac{{x}−\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{x}}\Rightarrow{N}=\:\frac{\frac{\mathrm{1}}{{x}}}{\frac{\mathrm{1}}{{x}}−\mathrm{1}}=\:\frac{\mathrm{1}}{{x}\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{1}−{x}}…

Question Number 161241 by help last updated on 14/Dec/21 Answered by mr W last updated on 15/Dec/21 $${let}\:{y}={xt} \\ $$$${y}'={t}+{x}\frac{{dt}}{{dx}}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}}−{t}=−\frac{\mathrm{1}}{{t}} \\…

Question Number 30160 by naka3546 last updated on 17/Feb/18 Commented by ajfour last updated on 18/Feb/18 $${is}\:{it}\:{right}\:? \\ $$ Commented by ajfour last updated on…

Question Number 30148 by .none. last updated on 17/Feb/18 $${how}\:{to}\:{solve}\:\sqrt{\left({c}−\sqrt{−{c}}\right)^{\mathrm{2}} }+\sqrt{\left({c}+\mathrm{6}−\sqrt{−{c}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}<{a}<\mathrm{3},\:\:−\mathrm{4}<{b}<−\mathrm{3},\:\:{c}=\frac{{a}+{b}}{\mathrm{2}} \\ $$ Answered by mrW2 last updated on 17/Feb/18 $$−\mathrm{2}<{a}+{b}<\mathrm{0} \\…