5/y+5-9/y-5=10/y^2-25

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Solution for 5/y+5-9/y-5=10/y^2-25 equation:


D( y )

y = 0

y^2 = 0

y = 0

y = 0

y^2 = 0

y^2 = 0

1*y^2 = 0 // : 1

y^2 = 0

y = 0

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

5/y-(9/y)-5+5 = 10/(y^2)-25 // - 10/(y^2)-25

5/y-(9/y)-(10/(y^2))-5+5+25 = 0

5/y-9*y^-1-10*y^-2-5+5+25 = 0

25-4*y^-1-10*y^-2 = 0

t_1 = y^-1

25-10*t_1^2-4*t_1^1 = 0

25-10*t_1^2-4*t_1 = 0

DELTA = (-4)^2-(-10*4*25)

DELTA = 1016

DELTA > 0

t_1 = (1016^(1/2)+4)/(-10*2) or t_1 = (4-1016^(1/2))/(-10*2)

t_1 = (2*254^(1/2)+4)/(-20) or t_1 = (4-2*254^(1/2))/(-20)

t_1 = (2*254^(1/2)+4)/(-20)

y^-1-((2*254^(1/2)+4)/(-20)) = 0

1*y^-1 = (2*254^(1/2)+4)/(-20) // : 1

y^-1 = (2*254^(1/2)+4)/(-20)

-1 < 0

1/(y^1) = (2*254^(1/2)+4)/(-20) // * y^1

1 = ((2*254^(1/2)+4)/(-20))*y^1 // : (2*254^(1/2)+4)/(-20)

-20*(2*254^(1/2)+4)^-1 = y^1

y = -20*(2*254^(1/2)+4)^-1

t_1 = (4-2*254^(1/2))/(-20)

y^-1-((4-2*254^(1/2))/(-20)) = 0

1*y^-1 = (4-2*254^(1/2))/(-20) // : 1

y^-1 = (4-2*254^(1/2))/(-20)

-1 < 0

1/(y^1) = (4-2*254^(1/2))/(-20) // * y^1

1 = ((4-2*254^(1/2))/(-20))*y^1 // : (4-2*254^(1/2))/(-20)

-20*(4-2*254^(1/2))^-1 = y^1

y = -20*(4-2*254^(1/2))^-1

y in { -20*(2*254^(1/2)+4)^-1, -20*(4-2*254^(1/2))^-1 }

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