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avatar+172 

Find:

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

 May 8, 2020
 #1
avatar+172 
0

Nevermind found i!. It's \(\sqrt{2}\)

.
 May 8, 2020
 #2
avatar+24951 
+1

Find:

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}.\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}} \\ x-1 &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ x-1 &=& \cfrac{1}{2 + x-1} \\ (x-1)(2 + x-1) &=& 1 \\ x+x^2-x-2-x+1 &=& 1 \\ x^2-2 &=& 0 \\ x^2 &=& 2 \\ \mathbf{ x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}\)

 

laugh

 May 8, 2020

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