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