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avatar+272 

3x=x3x

it solves out fine manually and all its just interesting.

 Apr 26, 2015

Best Answer 

 #4
avatar+118696 
+10

Lets just proove this TheJonyMyster

 

\\\sqrt[n]{x}=\sqrt[(n-1)]{\frac{x}{\sqrt[n]{x}}}\\\\ RHS=\sqrt[(n-1)]{x^{1-\frac{1}{n}}}\\\\ RHS=\sqrt[(n-1)]{x^{\frac{n-1}{n}}}\\\\ RHS=x^{\frac{n-1}{n}}\right)^{\frac{1}{n-1}}\\\\ RHS=x^{(1/n)}\\\\ RHS=\sqrt[n]{x}\\\\ RHS=LHS    

 Apr 28, 2015
 #1
avatar+33654 
+10

It's true as long as x is greater than zero.

.

 Apr 26, 2015
 #2
avatar+118696 
+5

Yes TheJonyMyster, it is cool how roots work :)

 Apr 26, 2015
 #3
avatar+272 
+5

yeah, positive x cause of imaginary numbers. also it works for any nx=(n1)xnx

 Apr 26, 2015
 #4
avatar+118696 
+10
Best Answer

Lets just proove this TheJonyMyster

 

\\\sqrt[n]{x}=\sqrt[(n-1)]{\frac{x}{\sqrt[n]{x}}}\\\\ RHS=\sqrt[(n-1)]{x^{1-\frac{1}{n}}}\\\\ RHS=\sqrt[(n-1)]{x^{\frac{n-1}{n}}}\\\\ RHS=x^{\frac{n-1}{n}}\right)^{\frac{1}{n-1}}\\\\ RHS=x^{(1/n)}\\\\ RHS=\sqrt[n]{x}\\\\ RHS=LHS    

Melody Apr 28, 2015
 #5
avatar+130466 
0

Nice proof,  Melody......!!!!

 

  

 Apr 28, 2015
 #6
avatar+118696 
0

thanks Chris :)

 Apr 28, 2015

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