Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 61241 by Tawa1 last updated on 30/May/19

Answered by meme last updated on 30/May/19

$$thelargerisAbecause{4}^{2015}-2^{2015}+1<4^{2015}+2^{2015}+1$$$$$$

Commented by Tawa1 last updated on 31/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by meme last updated on 30/May/19

$$thelargerisAbecause{4}^{2015}-2^{2015}+1<4^{2015}+2^{2015}+1$$

Answered by MJS last updated on 31/May/19

$$\frac{2^x-1}{2^{2x}-2^x+1}<?>\frac{2^x+1}{2^{2x}+2^x+1}$$$$\frac{a-1}{a^2-a+1}<?>\frac{a+1}{a^2+a+1}$$$$\frac{1}{a+\frac{1}{a-1}}<?>\frac{1}{a+\frac{1}{a+1}}$$$$\mathrm{now}$$$$a-1<a+1\Rightarrow \frac{1}{a-1}>\frac{1}{a+1}\Rightarrow a+\frac{1}{a-1}>a+\frac{1}{a+1}\Rightarrow$$$$\Rightarrow \frac{1}{a+\frac{1}{a-1}}<\frac{1}{a+\frac{1}{a+1}}\Rightarrow$$$$\Rightarrow A<B$$

Commented by Tawa1 last updated on 31/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by MJS last updated on 31/May/19

$$\frac{2^{2015}-1}{4^{2015}-2^{2015}+1}-\frac{2^{2015}+1}{4^{2015}+2^{2015}+1}=$$$$=\frac{(2^{2015}-1)(4^{2015}+2^{2015}+1)-(2^{2015}+1)(4^{2015}-2^{2015}+1)}{(4^{2015}-2^{2015}+1)(4^{2015}+2^{2015}+1)}=$$$$=\frac{-2}{{16}^{2015}+4^{2015}+1}<0\Rightarrow A<B$$

Commented by Tawa1 last updated on 31/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by Rasheed.Sindhi last updated on 31/May/19

$$\mathrm{A}=\frac{2^{2015}-1}{{\left(2^{2015}\right)}^2-2^{2015}+1}\times \frac{2^{2015}+1}{2^{2015}+1}$$$$=\frac{{\left(2^{2015}\right)}^2-1}{{\left(2^{2015}\right)}^3+1}$$$$\mathrm{B}=\frac{2^{2015}+1}{{\left(2^{2015}\right)}^2+2^{2015}+1}\times \frac{2^{2015}-1}{2^{2015}-1}$$$$=\frac{{\left(2^{2015}\right)}^2-1}{{\left(2^{2015}\right)}^3-1}$$$$$$$$\mathrm{N}\left(\mathrm{A}\right)=\mathrm{N}\left(\mathrm{B}\right)$$$$\mathrm{\&}$$$$\mathrm{D}\left(\mathrm{A}\right)>\mathrm{D}\left(\mathrm{B}\right)$$$$\therefore \mathrm{A}<\mathrm{B}$$

Commented by Tawa1 last updated on 31/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com