(1-2x)/(x+2)+20/(x+2)(x-2)=(2x+1)/(2-x)

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Solution for (1-2x)/(x+2)+20/(x+2)(x-2)=(2x+1)/(2-x) equation: 


D( x )

2-x = 0

x+2 = 0

2-x = 0

2-x = 0

2-x = 0 // - 2

-x = -2 // * -1

x = 2

x+2 = 0

x+2 = 0

x+2 = 0 // - 2

x = -2

x in (-oo:-2) U (-2:2) U (2:+oo)

(1-(2*x))/(x+2)+(20/(x+2))*(x-2) = (2*x+1)/(2-x) // - (2*x+1)/(2-x)

(1-(2*x))/(x+2)+(20/(x+2))*(x-2)-((2*x+1)/(2-x)) = 0

(1-2*x)/(x+2)+(20/(x+2))*(x-2)-((2*x+1)/(2-x)) = 0

(1-2*x)/(x+2)+(20*(x-2))/(x+2)+(-1*(2*x+1))/(2-x) = 0

((1-2*x)*(2-x))/((x+2)*(2-x))+(20*(x-2)*(2-x))/((x+2)*(2-x))+(-1*(2*x+1)*(x+2))/((x+2)*(2-x)) = 0

(1-2*x)*(2-x)+20*(x-2)*(2-x)-1*(2*x+1)*(x+2) = 0

75*x-18*x^2-2*x^2-5*x-78-2 = 0

70*x-20*x^2-80 = 0

70*x-20*x^2-80 = 0

10*(7*x-2*x^2-8) = 0

7*x-2*x^2-8 = 0

DELTA = 7^2-(-8*(-2)*4)

DELTA = -15

DELTA < 0

10 = 0

10/((x+2)*(2-x)) = 0

10/((x+2)*(2-x)) = 0 // * (x+2)*(2-x)

10 = 0

x belongs to the empty set

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