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}\:\:\:\: \\ $$$$\begin{cases}{\frac{{y}+{z}}{{x}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}+\mathrm{1}}\\{\frac{{x}+{z}}{{y}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}{y}−\mathrm{1}}+\mathrm{1}}\\{\frac{{x}+{z}}{{z}}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{5}{z}−\mathrm{1}}+\mathrm{1}}\end{cases}\:\:\:\: \\ $$$$\begin{cases}{\frac{{x}+{y}+{z}}{{x}}=\frac{\mathrm{2}{x}}{\mathrm{2}{x}−\mathrm{1}}}\\{\frac{{x}+{y}+{z}}{{y}}=\frac{\mathrm{3}{y}}{\mathrm{3}{y}−\mathrm{1}}}\\{\frac{{x}+{y}+{z}}{{z}}=\frac{\mathrm{5}{z}}{\mathrm{5}{z}−\mathrm{1}}}\end{cases}\:\:\:\: \\ $$$$\begin{cases}{{x}+{y}+{z}=\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{x}−\mathrm{1}}}\\{{x}+{y}+{z}=\frac{\mathrm{3}{y}^{\mathrm{2}} }{\mathrm{3}{y}−\mathrm{1}}\:\:\:\:\:\:\:\:}\\{{x}+{y}+{z}=\frac{\mathrm{5}{z}^{\mathrm{2}}…
Question Number 40355 by jasno91 last updated on 20/Jul/18 Answered by $@ty@m last updated on 20/Jul/18 Commented by $@ty@m last updated on 20/Jul/18 Terms of…
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Question Number 40344 by behi83417@gmail.com last updated on 20/Jul/18 Commented by MJS last updated on 20/Jul/18 $$\mathrm{maybe}\:\mathrm{we}\:\mathrm{can}\:\mathrm{show}\:{p}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Commented by behi83417@gmail.com last updated on…
Question Number 171403 by mnjuly1970 last updated on 14/Jun/22 $$\:\:{a}\:,\:{b}\:,\:{c}\:\in\mathbb{R}\:\:{and}\:\:{a}\neq{b}\neq{c} \\ $$$$\:\:{and}\:\:\:\frac{{a}}{{b}+{c}}\:,\:\frac{{b}}{{a}+{c}}\:,\:\frac{{c}}{{a}+{b}}\:\:\:{are}\:{three}\:{consecutive}\:{terms}\:{of}\:{AP} \\ $$$$\:\:\left({Arithmetic}\:{progression}\:\right) \\ $$$$\:\:{find}\:−:\:\:\:\frac{{b}^{\:\mathrm{2}} −{a}^{\:\mathrm{2}} }{{c}^{\:\mathrm{2}} −{b}^{\:\mathrm{2}} } \\ $$ Answered by Rasheed.Sindhi…
Question Number 171397 by mathlove last updated on 14/Jun/22 $${ax}+{by}+{cz}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${x}+{y}+{z}=\mathrm{18} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=? \\ $$ Commented by mr W last updated on 14/Jun/22 $$\mathrm{2}\:{equations}\:{for}\:\mathrm{6}\:{variables},\:…