Question Number 208819 by Ismoiljon_008 last updated on 23/Jun/24
Answered by Ismoiljon_008 last updated on 23/Jun/24
$${help}\:{please} \\ $$
Answered by A5T last updated on 23/Jun/24
![4a^4 +1=(2a^2 )^2 +1^2 =(2a^2 +1)^2 −(2a)^2 =(2a^2 −2a+1)_(a) (2a^2 +1+2a)_(b) 4(a+1)^4 +1=[2(a+1)^2 ]^2 +1=[2(a+1)^2 +1]^2 −4(a+1)^2 =[2(a+1)^2 +1−2(a+1)][2(a+1)^2 +1+2(a+1)] =[2a^2 +2a+1]_(b) [2a^2 +6a+5]_(c) So, 4a^4 +1=ab⇒4(a+1)^4 +1=bc Question⇒((ab×cd×ef×gh×ij)/(bc×de×fg×hi×jk)) =(a/k)=((2(1)^2 −2(1)+1)/(2(9)^2 +6(9)+5))=(1/(221))](https://www.tinkutara.com/question/Q208824.png)
$$\mathrm{4}{a}^{\mathrm{4}} +\mathrm{1}=\left(\mathrm{2}{a}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} =\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}\right)^{\mathrm{2}} \\ $$$$=\underset{{a}} {\underbrace{\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}\right)}}\:\underset{{b}} {\underbrace{\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{a}\right)}} \\ $$$$\mathrm{4}\left({a}+\mathrm{1}\right)^{\mathrm{4}} +\mathrm{1}=\left[\mathrm{2}\left({a}+\mathrm{1}\right)^{\mathrm{2}} \right]^{\mathrm{2}} +\mathrm{1}=\left[\mathrm{2}\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right]^{\mathrm{2}} −\mathrm{4}\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\left[\mathrm{2}\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\left({a}+\mathrm{1}\right)\right]\left[\mathrm{2}\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}+\mathrm{2}\left({a}+\mathrm{1}\right)\right] \\ $$$$=\underset{{b}} {\underbrace{\left[\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}\right]}}\:\underset{{c}} {\underbrace{\left[\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{5}\right]}} \\ $$$${So},\:\mathrm{4}{a}^{\mathrm{4}} +\mathrm{1}={ab}\Rightarrow\mathrm{4}\left({a}+\mathrm{1}\right)^{\mathrm{4}} +\mathrm{1}={bc} \\ $$$${Question}\Rightarrow\frac{{ab}×{cd}×{ef}×{gh}×{ij}}{{bc}×{de}×{fg}×{hi}×{jk}} \\ $$$$=\frac{{a}}{{k}}=\frac{\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}\right)+\mathrm{1}}{\mathrm{2}\left(\mathrm{9}\right)^{\mathrm{2}} +\mathrm{6}\left(\mathrm{9}\right)+\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{221}} \\ $$
Commented by Ismoiljon_008 last updated on 24/Jun/24
$${thank}\:{you}\:{very}\:{much} \\ $$