Menu Close

5-6-1-8-1-1-6-1-4-1-1-6-1-1-




Question Number 146896 by mathdanisur last updated on 16/Jul/21
(5/( (6)^(1/8)  + 1)) ∙ (1/( (6)^(1/4)  + 1)) ∙ (1/( (√6) + 1)) + 1 = ?
$$\frac{\mathrm{5}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\:+\:\mathrm{1}}\:+\:\mathrm{1}\:=\:? \\ $$
Answered by EDWIN88 last updated on 16/Jul/21
 What the value of     (5/( (6)^(1/8)  +1)) ×(1/( (6)^(1/4)  +1)) ×(1/( (√6) +1)) + 1 .   Solution :   let (6)^(1/8)  = u → { (((6)^(1/4)  = u^2 )),(((√6) = u^4  )) :}  T = (5/( (6)^(1/8)  +1)) × (1/( (6)^(1/4) +1)) × (1/( (√6)+1)) +1  T=(5/(u+1))×(1/(u^2 +1))×(1/(u^4 +1))+1  T=(5/((u+1)(u^2 +1)(u^4 +1))) +1  T=(5/((u+1)(u^6 +u^4 +u^2 +1))) +1  T= (5/(u^7 +u^6 +u^5 +u^4 +u^3 +u^2 +u+1)) +1  T=((5(u−1))/(u^8 −1)) +1  T=((5((6)^(1/8) −1))/(6−1)) +1 = (6)^(1/8)
$$\:\mathrm{What}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\frac{\mathrm{5}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\mathrm{1}}\:×\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\mathrm{1}}\:×\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\:+\mathrm{1}}\:+\:\mathrm{1}\:. \\ $$$$\:\mathrm{Solution}\::\: \\ $$$$\mathrm{let}\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:=\:\mathrm{u}\:\rightarrow\begin{cases}{\sqrt[{\mathrm{4}}]{\mathrm{6}}\:=\:\mathrm{u}^{\mathrm{2}} }\\{\sqrt{\mathrm{6}}\:=\:\mathrm{u}^{\mathrm{4}} \:}\end{cases} \\ $$$$\mathrm{T}\:=\:\frac{\mathrm{5}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\mathrm{1}}\:×\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}+\mathrm{1}}\:×\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}+\mathrm{1}}\:+\mathrm{1} \\ $$$$\mathrm{T}=\frac{\mathrm{5}}{\mathrm{u}+\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{4}} +\mathrm{1}}+\mathrm{1} \\ $$$$\mathrm{T}=\frac{\mathrm{5}}{\left(\mathrm{u}+\mathrm{1}\right)\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{u}^{\mathrm{4}} +\mathrm{1}\right)}\:+\mathrm{1} \\ $$$$\mathrm{T}=\frac{\mathrm{5}}{\left(\mathrm{u}+\mathrm{1}\right)\left(\mathrm{u}^{\mathrm{6}} +\mathrm{u}^{\mathrm{4}} +\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}\:+\mathrm{1} \\ $$$$\mathrm{T}=\:\frac{\mathrm{5}}{\mathrm{u}^{\mathrm{7}} +\mathrm{u}^{\mathrm{6}} +\mathrm{u}^{\mathrm{5}} +\mathrm{u}^{\mathrm{4}} +\mathrm{u}^{\mathrm{3}} +\mathrm{u}^{\mathrm{2}} +\mathrm{u}+\mathrm{1}}\:+\mathrm{1} \\ $$$$\mathrm{T}=\frac{\mathrm{5}\left(\mathrm{u}−\mathrm{1}\right)}{\mathrm{u}^{\mathrm{8}} −\mathrm{1}}\:+\mathrm{1} \\ $$$$\mathrm{T}=\frac{\mathrm{5}\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}−\mathrm{1}\right)}{\mathrm{6}−\mathrm{1}}\:+\mathrm{1}\:=\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\: \\ $$
Commented by mathdanisur last updated on 16/Jul/21
thank you Ser
$${thank}\:{you}\:{Ser} \\ $$
Answered by liberty last updated on 16/Jul/21
 (5/( (6)^(1/8) +1)) ×(((6)^(1/8) −1)/( (6)^(1/8) −1))×(1/( (6)^(1/4) +1))×(1/( (√6)+1))+1   =((5((6)^(1/8) −1))/( (6)^(1/4) −1))×(1/( (6)^(1/4) +1))×(1/( (√6)+1))+1  = ((5((6)^(1/8) −1))/( (√6)−1))×(1/( (√6)+1)) +1   = ((5((6)^(1/8) −1))/(6−1)) + 1   = (6)^(1/8) −1+1= (6)^(1/8)  .
$$\:\frac{\mathrm{5}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}+\mathrm{1}}\:×\frac{\sqrt[{\mathrm{8}}]{\mathrm{6}}−\mathrm{1}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}−\mathrm{1}}×\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}+\mathrm{1}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}+\mathrm{1}}+\mathrm{1}\: \\ $$$$=\frac{\mathrm{5}\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}−\mathrm{1}\right)}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}−\mathrm{1}}×\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}+\mathrm{1}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}+\mathrm{1}}+\mathrm{1} \\ $$$$=\:\frac{\mathrm{5}\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{6}}−\mathrm{1}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}+\mathrm{1}}\:+\mathrm{1}\: \\ $$$$=\:\frac{\mathrm{5}\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}−\mathrm{1}\right)}{\mathrm{6}−\mathrm{1}}\:+\:\mathrm{1}\: \\ $$$$=\:\sqrt[{\mathrm{8}}]{\mathrm{6}}−\mathrm{1}+\mathrm{1}=\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:. \\ $$
Commented by mathdanisur last updated on 16/Jul/21
thank you Ser
$${thank}\:{you}\:{Ser} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *