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avatar+257 

[x^(1/(a-b)*(1/a-c)]*[x^(1/b-a)*(1/b-c)]*[x^(1/c-a)*(1/c-b)]

 Mar 4, 2016

Best Answer 

 #5
avatar+15075 
+15

Hello AaratrikRoy!

 

[x^(1/(a-b)*(1/a-c)]*[x^(1/b-a)*(1/b-c)]*[x^(1/c-a)*(1/c-b)]

 

[x^(1 / (a - b) * (1/a - c)] * [x^(1/b - a) * (1/b - c)] * [x^(1/c - a) * (1/c - b)]

 

= [x1/(a-b) * (-c + 1/a)] * [x -a+1/b * (-c + 1/b)] * [x -a+1/c * (-b + 1/c)]       ????

 

Have I correctly interpreted?
Complete please the red marked clip!
Greetings asinus :- )

 laugh!

 Mar 4, 2016
 #1
avatar+2499 
+4
 Mar 4, 2016
 #2
avatar+257 
0

I use it,but it is not working

AaratrikRoy  Mar 4, 2016
 #3
avatar+2499 
+4

use only "( )"  not "[ ]"

Solveit  Mar 4, 2016
 #4
avatar+26396 
+15

[x^(1/(a-b)*(1/a-c)]*[x^(1/b-a)*(1/b-c)]*[x^(1/c-a)*(1/c-b)]

 

one ")" missing

 

laugh

 Mar 4, 2016
 #5
avatar+15075 
+15
Best Answer

Hello AaratrikRoy!

 

[x^(1/(a-b)*(1/a-c)]*[x^(1/b-a)*(1/b-c)]*[x^(1/c-a)*(1/c-b)]

 

[x^(1 / (a - b) * (1/a - c)] * [x^(1/b - a) * (1/b - c)] * [x^(1/c - a) * (1/c - b)]

 

= [x1/(a-b) * (-c + 1/a)] * [x -a+1/b * (-c + 1/b)] * [x -a+1/c * (-b + 1/c)]       ????

 

Have I correctly interpreted?
Complete please the red marked clip!
Greetings asinus :- )

 laugh!

asinus Mar 4, 2016
 #6
avatar+257 
0

(x^(1/(a-b)*(1/a-c))*(x^(1/b-a)*(1/b-c))*(x^(1/c-a)*(1/c-b))

AaratrikRoy  Mar 6, 2016

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