4x/x^2-4=8/x^2-4-4/x+2

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Solution for 4x/x^2-4=8/x^2-4-4/x+2 equation: 


D( x )

x = 0

x^2 = 0

x = 0

x = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

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

(4*x)/(x^2)-4 = 8/(x^2)-(4/x)-4+2 // - 8/(x^2)-(4/x)-4+2

(4*x)/(x^2)+4/x-(8/(x^2))-4-2+4 = 0

(4*x)/(x^2)+4/x-8*x^-2-4-2+4 = 0

8*x^-1-8*x^-2-2 = 0

t_1 = x^-1

8*t_1^1-8*t_1^2-2 = 0

8*t_1-8*t_1^2-2 = 0

DELTA = 8^2-(-8*(-2)*4)

DELTA = 0

t_1 = -8/(-8*2)

t_1 = 1/2 or t_1 = 1/2

t_1 = 1/2

x^-1-1/2 = 0

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

x^-1 = 1/2

-1 < 0

1/(x^1) = 1/2 // * x^1

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

2 = x^1

x = 2

t_1 = 1/2

x^-1-1/2 = 0

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

x^-1 = 1/2

-1 < 0

1/(x^1) = 1/2 // * x^1

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

2 = x^1

x = 2

x in { 2, 2 }

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