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1/2(x)(x+2)+(2x)(2x)+(2x+2)(x+2)=136
We move all terms to the left:
1/2(x)(x+2)+(2x)(2x)+(2x+2)(x+2)-(136)=0
Domain of the equation: 2x(x+2)!=0We multiply parentheses ..
x∈R
(+2x^2+4x+2x+4)+1/2x(x+2)+2x2x-136=0
We multiply all the terms by the denominator
((+2x^2+4x+2x+4))*2x(x+2)+2x2x*2x(x+2)-136*2x(x+2)+1=0
We calculate terms in parentheses: +((+2x^2+4x+2x+4))*2x(x+2), so:Wy multiply elements
(+2x^2+4x+2x+4))*2x(x+2
((+2x^2+4x+2x+4))*2x(x+2)+4x^3(x-272x(x+1=0
We calculate terms in parentheses: +((+2x^2+4x+2x+4))*2x(x+2), so:We do not support expression: x^3(x
(+2x^2+4x+2x+4))*2x(x+2
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