(2/x+1)+(3/X-1)=(5/x^2-1)

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Solution for (2/x+1)+(3/X-1)=(5/x^2-1) 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)

3/X+2/x-1+1 = 5/(x^2)-1 // - 5/(x^2)-1

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

3/X+2/x-5*x^-2-1+1+1 = 0

3*X^-1+2*x^-1-5*x^-2+1 = 0

t_1 = x^-1

3*X^-1-5*t_1^2+2*t_1^1+1 = 0

3*X^-1-5*t_1^2+2*t_1+1 = 0

DELTA = 2^2-(-5*4*(3*X^-1+1))

DELTA = 20*(3*X^-1+1)+4

20*(3*X^-1+1)+4 = 0

20*(3*X^-1+1)+4 = 0

60*X^-1+24 = 0

60*X^-1+24 = 0

60*X^-1 = -24 // : 60

X^-1 = -2/5

-1 < 0

1/(X^1) = -2/5 // * X^1

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

-5/2 = X^1

X = -5/2

DELTA = 0 <=> t_2 = -5/2

t_1 = -2/(-5*2) i X = -5/2

t_1 = 1/5 i X = -5/2

( t_1 = ((20*(3*X^-1+1)+4)^(1/2)-2)/(-5*2) or t_1 = (-(20*(3*X^-1+1)+4)^(1/2)-2)/(-5*2) ) i X > -5/2

( t_1 = ((20*(3*X^-1+1)+4)^(1/2)-2)/(-10) or t_1 = ((20*(3*X^-1+1)+4)^(1/2)+2)/10 ) i X > -5/2

X+5/2 > 0

X+5/2 > 0 // - 5/2

X > -5/2

t_1 = 1/5

x^-1-1/5 = 0

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

x^-1 = 1/5

-1 < 0

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

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

5 = x^1

x = 5

t_1 = ((20*(3*X^-1+1)+4)^(1/2)-2)/(-10)

t_3 = -(((20*(3*X^-1+1)+4)^(1/2)-2)/(-10))

t_3+x^-1 = 0

1*x^-1 = -t_3 // : 1

x^-1 = -t_3

-1 < 0

1/(x^1) = -t_3 // * x^1

1 = -t_3*x^1 // : -t_3

-t_3^-1 = x^1

x = -t_3^-1

t_1 = ((20*(3*X^-1+1)+4)^(1/2)+2)/10

t_4 = -(((20*(3*X^-1+1)+4)^(1/2)+2)/10)

t_4+x^-1 = 0

1*x^-1 = -t_4 // : 1

x^-1 = -t_4

-1 < 0

1/(x^1) = -t_4 // * x^1

1 = -t_4*x^1 // : -t_4

-t_4^-1 = x^1

x = -t_4^-1

x in { 5, -(-(((20*(3*X^-1+1)+4)^(1/2)-2)/(-10)))^-1, -(-(((20*(3*X^-1+1)+4)^(1/2)+2)/10))^-1 }

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