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(1)/(5-x)=1/(8x)
We move all terms to the left:
(1)/(5-x)-(1/(8x))=0
Domain of the equation: (5-x)!=0
We move all terms containing x to the left, all other terms to the right
-x!=-5
x!=-5/-1
x!=+5
x∈R
Domain of the equation: 8x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/(-1x+5)-(+1/8x)=0
We get rid of parentheses
1/(-1x+5)-1/8x=0
We calculate fractions
8x/(-8x^2+40x)+(-1*(-1x+5))/(-8x^2+40x)=0
We calculate terms in parentheses: +(-1*(-1x+5))/(-8x^2+40x), so:We multiply all the terms by the denominator
-1*(-1x+5))/(-8x^2+40x
determiningTheFunctionDomain -8x^2-1*(-1x+5))/(+40x
We add all the numbers together, and all the variables
-8x^2+40x-1*(-1x+5))/(
We multiply all the terms by the denominator
-8x^2*(+40x*(-1*(-1x+5))
Back to the equation:
+(-8x^2*(+40x*(-1*(-1x+5)))
((-8x^2*(+40x*(-1*(-1x+5))))*(-8x^2+40x)+8x=0
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