1/(x-3)^2-4/(x-3)-5=0

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Solution for 1/(x-3)^2-4/(x-3)-5=0 equation: 



1/(x-3)^2-4/(x-3)-5=0



Domain of the equation: (x-3)^2!=0

x∈R





Domain of the equation: (x-3)!=0

We move all terms containing x to the left, all other terms to the right

x!=3

x∈R



We calculate fractions


(1*(x-3))/((x-3)^2*(x-3))+(-4*(x-3)^2)/((x-3)^2*(x-3))-5=0



We calculate terms in parentheses: +(1*(x-3))/((x-3)^2*(x-3)), so:

1*(x-3))/((x-3)^2*(x-3)

We multiply all the terms by the denominator


1*(x-3))

Back to the equation:

+(1*(x-3)))





We calculate terms in parentheses: +(-4*(x-3)^2)/((x-3)^2*(x-3)), so:

-4*(x-3)^2)/((x-3)^2*(x-3)

We multiply all the terms by the denominator


-4*(x-3)^2)

We multiply parentheses

-4x-

We add all the numbers together, and all the variables

-4x

Back to the equation:

+(-4x)

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