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1/4(x-1)+2/5(x-5)=1/5(x^2-53)
We move all terms to the left:
1/4(x-1)+2/5(x-5)-(1/5(x^2-53))=0
Domain of the equation: 4(x-1)!=0
x∈R
Domain of the equation: 5(x-5)!=0
x∈R
Domain of the equation: 5(x^2-53))!=0We calculate fractions
x∈R
(-20x^2x/(4(x-1)*5(x-5)*5(x^2-53)))+(25x^2x/(4(x-1)*5(x-5)*5(x^2-53)))+(40x^2x/(4(x-1)*5(x-5)*5(x^2-53)))=0
We calculate terms in parentheses: +(-20x^2x/(4(x-1)*5(x-5)*5(x^2-53))), so:
-20x^2x/(4(x-1)*5(x-5)*5(x^2-53))
We multiply all the terms by the denominator
-20x^2x
Back to the equation:
+(-20x^2x)
We calculate terms in parentheses: +(25x^2x/(4(x-1)*5(x-5)*5(x^2-53))), so:
25x^2x/(4(x-1)*5(x-5)*5(x^2-53))
We multiply all the terms by the denominator
25x^2x
Back to the equation:
+(25x^2x)
We calculate terms in parentheses: +(40x^2x/(4(x-1)*5(x-5)*5(x^2-53))), so:We get rid of parentheses
40x^2x/(4(x-1)*5(x-5)*5(x^2-53))
We multiply all the terms by the denominator
40x^2x
Back to the equation:
+(40x^2x)
-20x^2x+25x^2x+40x^2x=0
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