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5(2x+1)^2/(4x+5)=5x-1
We move all terms to the left:
5(2x+1)^2/(4x+5)-(5x-1)=0
Domain of the equation: (4x+5)!=0We get rid of parentheses
We move all terms containing x to the left, all other terms to the right
4x!=-5
x!=-5/4
x!=-1+1/4
x∈R
5(2x+1)^2/(4x+5)-5x+1=0
We multiply all the terms by the denominator
5(2x+1)^2-5x*(4x+5)+1*(4x+5)=0
We multiply parentheses
-20x^2+5(2x+1)^2-25x+1*(4x+5)=0
We add all the numbers together, and all the variables
-20x^2-25x+5(2x+1)^2+1*(4x+5)=0
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