1=(6-3x/x^2+x)+(3/2x+2)

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Solution for 1=(6-3x/x^2+x)+(3/2x+2) equation:


D( x )

x^2 = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

x in (-oo:0) U (0:+oo)

1 = x+(3/2)*x-((3*x)/(x^2))+2+6 // - x+(3/2)*x-((3*x)/(x^2))+2+6

(3*x)/(x^2)-x-((3/2)*x)-6-2+1 = 0

(-3/2)*x-x+(3*x)/(x^2)-6-2+1 = 0

3*x^-1-5/2*x^1-7*x^0 = 0

(3*x^0-5/2*x^2-7*x^1)/(x^1) = 0 // * x^2

x^1*(3*x^0-5/2*x^2-7*x^1) = 0

x^1

(-5/2)*x^2-7*x+3 = 0

(-5/2)*x^2-7*x+3 = 0

DELTA = (-7)^2-(3*4*(-5/2))

DELTA = 79

DELTA > 0

x = (79^(1/2)+7)/(2*(-5/2)) or x = (7-79^(1/2))/(2*(-5/2))

x = (79^(1/2)+7)/(-5) or x = (7-79^(1/2))/(-5)

x in { (79^(1/2)+7)/(-5), (7-79^(1/2))/(-5)}

x in { (79^(1/2)+7)/(-5), (7-79^(1/2))/(-5) }

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