4n/n^2=5/n-1/n^2

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Solution for 4n/n^2=5/n-1/n^2 equation:


D( n )

n = 0

n^2 = 0

n = 0

n = 0

n^2 = 0

n^2 = 0

1*n^2 = 0 // : 1

n^2 = 0

n = 0

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

(4*n)/(n^2) = 5/n-(1/(n^2)) // - 5/n-(1/(n^2))

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

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

n^-2-n^-1 = 0

t_1 = n^-1

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

t_1^2-t_1 = 0

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

DELTA = 1

DELTA > 0

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

t_1 = 1 or t_1 = 0

t_1 = 0

n^-1+0 = 0

n^-1 = 0

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

n^-1 = 0

n należy do O

t_1 = 1

n^-1-1 = 0

1*n^-1 = 1 // : 1

n^-1 = 1

-1 < 0

1/(n^1) = 1 // * n^1

1 = 1*n^1 // : 1

1 = n^1

n = 1

n = 1

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