(2/x-2)+(3/x+2)=(8/x^2/4)

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Solution for (2/x-2)+(3/x+2)=(8/x^2/4) 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)

2/x+3/x-2+2 = (8/(x^2))/4 // - (8/(x^2))/4

2/x+3/x-((8/(x^2))/4)-2+2 = 0

2/x+3/x-2*x^-2-2+2 = 0

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

t_1 = x^-1

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

5*t_1-2*t_1^2 = 0

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

DELTA = 25

DELTA > 0

t_1 = (25^(1/2)-5)/(-2*2) or t_1 = (-25^(1/2)-5)/(-2*2)

t_1 = 0 or t_1 = 5/2

t_1 = 0

x^-1+0 = 0

x^-1 = 0

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

x^-1 = 0

x naleu017Cy do O

t_1 = 5/2

x^-1-5/2 = 0

1*x^-1 = 5/2 // : 1

x^-1 = 5/2

-1 < 0

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

1 = 5/2*x^1 // : 5/2

2/5 = x^1

x = 2/5

x = 2/5

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