(5/(x-5/x))-(4/x)

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Solution for (5/(x-5/x))-(4/x) equation: 


D( x )

x-(5/x) = 0

x = 0

x-(5/x) = 0

x-(5/x) = 0

x-5*x^-1 = 0

1*x^1-5*x^-1 = 0

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

x^1*(1*x^2-5*x^0) = 0

x^1

x^2-5 = 0

x^2-5 = 0

DELTA = 0^2-(-5*1*4)

DELTA = 20

DELTA > 0

x = (20^(1/2)+0)/(1*2) or x = (0-20^(1/2))/(1*2)

x = 5^(1/2) or x = -5^(1/2)

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

x = 0

x = 0

x in (-oo:-5^(1/2)) U (-5^(1/2):0) U (0:5^(1/2)) U (5^(1/2):+oo)

5/(x-(5/x))-(4/x) = 0

5/(x-5*x^-1)-4*x^-1 = 0

5/(x-5*x^-1)-4/x = 0

x-5*x^-1 = 0

x-5*x^-1 = 0

x^-1*(x^2-5) = 0

1*x^2 = 5 // : 1

x^2 = 5

x^2 = 5 // ^ 1/2

abs(x) = 5^(1/2)

x = 5^(1/2) or x = -5^(1/2)

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

5/(x^-1*(x-5^(1/2))*(x+5^(1/2)))-4/x = 0

(5*x)/(x^-1*x*(x-5^(1/2))*(x+5^(1/2)))+(-4*x^-1*(x-5^(1/2))*(x+5^(1/2)))/(x^-1*x*(x-5^(1/2))*(x+5^(1/2))) = 0

5*x-4*x^-1*(x-5^(1/2))*(x+5^(1/2)) = 0

x+20*x^-1 = 0

x+20*x^-1 = 0

x^-1*(x^2+20) = 0

1*x^2 = -20 // : 1

x^2 = -20

1/x = 0

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

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

1/x = 0

1*x^-1 = 0 // : 1

x^-1 = 0

x naleu017Cy do O

x belongs to the empty set

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