Question Number 40469 by MrW3 last updated on 22/Jul/18 Commented by MrW3 last updated on 22/Jul/18 $${Find}\:{the}\:{volume}\:{of}\:{pyramid}. \\ $$ Commented by MJS last updated on…
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Question Number 105927 by bemath last updated on 01/Aug/20 $$\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}\: \\ $$ Commented by Rasheed.Sindhi last updated on 02/Aug/20 $$\left(\frac{{x}}{{x}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}\: \\…
Question Number 105929 by Study last updated on 01/Aug/20 Answered by Dwaipayan Shikari last updated on 01/Aug/20 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\left({x}^{{x}^{{x}} } \right)+{x}^{{x}} \left({logx}+\mathrm{1}\right)+\mathrm{4}{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}} \\ $$$$…
Question Number 171456 by mathlove last updated on 15/Jun/22 Commented by mathlove last updated on 16/Jun/22 $${solution}?? \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 40376 by behi83417@gmail.com last updated on 21/Jul/18 Commented by MrW3 last updated on 21/Jul/18 $$\left(\mathrm{1}\right) \\ $$$${let}\:{u}={x}−\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{u}+\mathrm{1}}+\frac{\mathrm{3}}{{u}−\mathrm{1}}=\mathrm{1}−\frac{\mathrm{2}}{{u}} \\ $$$$\frac{\mathrm{4}{u}+\mathrm{2}}{{u}^{\mathrm{2}} −\mathrm{1}}=\frac{{u}−\mathrm{2}}{{u}} \\…
Question Number 40375 by behi83417@gmail.com last updated on 21/Jul/18 Answered by MJS last updated on 21/Jul/18 $${xyz}\left({x}+{y}+{z}\right)=\mathrm{1}\:\Rightarrow\:{z}=−\frac{{x}+{y}}{\mathrm{2}}\pm\frac{\sqrt{{xy}\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}\sqrt{{xy}}}\:\Rightarrow \\ $$$$\Rightarrow\:{p}\left({x},\:{y},\:{z}\right)=\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}\right)}{{xy}} \\ $$$$\mathrm{after}\:\mathrm{drawing}\:\mathrm{lots}\:\mathrm{of}\:\mathrm{graphs}\:\mathrm{I}\:\mathrm{cane}\:\mathrm{to}\:\mathrm{the} \\…
Question Number 171443 by cortano1 last updated on 15/Jun/22 Answered by Rasheed.Sindhi last updated on 15/Jun/22 $$\begin{cases}{\frac{{y}+{z}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}}\\{\frac{{x}+{z}}{{y}}=\frac{\mathrm{1}}{\mathrm{3}{y}−\mathrm{1}}}\\{\frac{{x}+{z}}{{z}}=\frac{\mathrm{1}}{\mathrm{5}{z}−\mathrm{1}}}\end{cases}\:\:\:\: \\ $$$$\begin{cases}{\frac{{y}+{z}}{{x}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}+\mathrm{1}}\\{\frac{{x}+{z}}{{y}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}{y}−\mathrm{1}}+\mathrm{1}}\\{\frac{{x}+{z}}{{z}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{5}{z}−\mathrm{1}}+\mathrm{1}}\end{cases}\:\:\:\: \\ $$$$\begin{cases}{\frac{{x}+{y}+{z}}{{x}}=\frac{\mathrm{2}{x}}{\mathrm{2}{x}−\mathrm{1}}}\\{\frac{{x}+{y}+{z}}{{y}}=\frac{\mathrm{3}{y}}{\mathrm{3}{y}−\mathrm{1}}}\\{\frac{{x}+{y}+{z}}{{z}}=\frac{\mathrm{5}{z}}{\mathrm{5}{z}−\mathrm{1}}}\end{cases}\:\:\:\: \\ $$$$\begin{cases}{{x}+{y}+{z}=\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{x}−\mathrm{1}}}\\{{x}+{y}+{z}=\frac{\mathrm{3}{y}^{\mathrm{2}} }{\mathrm{3}{y}−\mathrm{1}}\:\:\:\:\:\:\:\:}\\{{x}+{y}+{z}=\frac{\mathrm{5}{z}^{\mathrm{2}}…