7/y^2-4-4/y+2=5/y-2

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Solution for 7/y^2-4-4/y+2=5/y-2 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)

7/(y^2)-(4/y)-4+2 = 5/y-2 // - 5/y-2

7/(y^2)-(4/y)-(5/y)-4+2+2 = 0

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

7*y^-2-9*y^-1 = 0

t_1 = y^-1

7*t_1^2-9*t_1^1 = 0

7*t_1^2-9*t_1 = 0

DELTA = (-9)^2-(0*4*7)

DELTA = 81

DELTA > 0

t_1 = (81^(1/2)+9)/(2*7) or t_1 = (9-81^(1/2))/(2*7)

t_1 = 9/7 or t_1 = 0

t_1 = 0

y^-1+0 = 0

y^-1 = 0

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

y^-1 = 0

y naleu017Cy do O

t_1 = 9/7

y^-1-9/7 = 0

1*y^-1 = 9/7 // : 1

y^-1 = 9/7

-1 < 0

1/(y^1) = 9/7 // * y^1

1 = 9/7*y^1 // : 9/7

7/9 = y^1

y = 7/9

y = 7/9

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