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9x/2(2x-5)(2x+5)+5x/x(2x-5)=1/x
We move all terms to the left:
9x/2(2x-5)(2x+5)+5x/x(2x-5)-(1/x)=0
Domain of the equation: 2(2x-5)(2x+5)!=0
x∈R
Domain of the equation: x(2x-5)!=0
x∈R
Domain of the equation: x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
9x/2(2x-5)(2x+5)+5x/x(2x-5)-(+1/x)=0
We use the square of the difference formula
4x^2+5x/x(2x-5)-(+1/x)-25=0
We get rid of parentheses
4x^2+5x/x(2x-5)-1/x-25=0
We calculate fractions
We do not support expression: x^3
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