Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 194971 by kapoorshah last updated on 21/Jul/23

Commented by Frix last updated on 21/Jul/23

$$\mathrm{I}\mathrm{think}{\left(abc\right)}^2=8$$

Commented by sniper237 last updated on 23/Jul/23

$$a+\frac{2}{b}=b+\frac{2}{c}\Rightarrow a-b=\frac{2}{c}-\frac{2}{b}\Rightarrow bc=\frac{2(b-c)}{a-b}$$$$Samelyac=\frac{2(c-a)}{b-c}andab=\frac{2(b-a)}{a-c}$$$$timingall{\left(abc\right)}^2=\left(ab\right)\left(ac\right)\left(bc\right)=8$$

Answered by Frix last updated on 21/Jul/23

$$a+\frac{2}{b}=b+\frac{2}{c}\wedge a+\frac{2}{b}=c+\frac{2}{a}$$$$c=\frac{2b}{ab-b^2+2}=\frac{a^{2}b+2a-2b}{ab}$$$$\mathrm{Transforming}\mathrm{and}\mathrm{factorizing}\mathrm{leads}\mathrm{to}$$$$(a-b)(a^2{b}^2+4ab-2b^2+4)=0$$$$\mathrm{But}a\ne b\Rightarrow$$$$b=-\frac{2(a\pm \sqrt{2})}{a^2-2^{(\ast )}}\wedge c=-\frac{2\mp a\sqrt{2}}{a}$$$$\Rightarrow a^2{b}^2{c}^2=8$$$${}^{(\ast )}\mathrm{Set}a=\sqrt{2}\mathrm{from}\mathrm{the}\mathrm{beginning}\Rightarrow \mathrm{we}\mathrm{get}$$$$b=-\frac{\sqrt{2}}{2}\wedge c=-2\sqrt{2}\mathrm{and}\mathrm{still}{a}^2{b}^2{c}^2=8$$

Commented by kapoorshah last updated on 21/Jul/23

$$nice$$$$thankyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com