(40/(x^2-16))-(5/(x-4))=(8/(x+4))

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Solution for (40/(x^2-16))-(5/(x-4))=(8/(x+4)) equation:


D( x )

x+4 = 0

x^2-16 = 0

x-4 = 0

x+4 = 0

x+4 = 0

x+4 = 0 // - 4

x = -4

x^2-16 = 0

x^2-16 = 0

1*x^2 = 16 // : 1

x^2 = 16

x^2 = 16 // ^ 1/2

abs(x) = 4

x = 4 or x = -4

x-4 = 0

x-4 = 0

x-4 = 0 // + 4

x = 4

x in (-oo:-4) U (-4:4) U (4:+oo)

40/(x^2-16)-(5/(x-4)) = 8/(x+4) // - 8/(x+4)

40/(x^2-16)-(5/(x-4))-(8/(x+4)) = 0

40/(x^2-16)-5*(x-4)^-1-8*(x+4)^-1 = 0

40/(x^2-16)-5/(x-4)-8/(x+4) = 0

(40*(x-4)*(x+4))/((x^2-16)*(x-4)*(x+4))+(-5*(x^2-16)*(x+4))/((x^2-16)*(x-4)*(x+4))+(-8*(x^2-16)*(x-4))/((x^2-16)*(x-4)*(x+4)) = 0

40*(x-4)*(x+4)-5*(x^2-16)*(x+4)-8*(x^2-16)*(x-4) = 0

20*x^2-5*x^3-8*x^3+32*x^2+80*x+128*x-512-320 = 0

52*x^2-13*x^3+208*x-832 = 0

52*x^2-13*x^3+208*x-832 = 0

13*(4*x^2-x^3+16*x-64) = 0

4*x^2-x^3+16*x-64 = 0

{ 1, -1, 2, -2, 4, -4, 8, -8, 16, -16, 32, -32, 64, -64 }

1

x = 1

4*x^2-x^3+16*x-64 = -45

1

-1

x = -1

4*x^2-x^3+16*x-64 = -75

-1

2

x = 2

4*x^2-x^3+16*x-64 = -24

2

-2

x = -2

4*x^2-x^3+16*x-64 = -72

-2

4

x = 4

4*x^2-x^3+16*x-64 = 0

4

x-4

16-x^2

4*x^2-x^3+16*x-64

x-4

x^3-4*x^2

16*x-64

64-16*x

0

16-x^2 = 0

DELTA = 0^2-(-1*4*16)

DELTA = 64

DELTA > 0

x = (64^(1/2)+0)/(-1*2) or x = (0-64^(1/2))/(-1*2)

x = -4 or x = 4

x in { -4, 4, 4}

13*(x+4)*(x-4)^2 = 0

(13*(x+4)*(x-4)^2)/((x^2-16)*(x-4)*(x+4)) = 0

(13*(x+4)*(x-4)^2)/((x^2-16)*(x-4)*(x+4)) = 0 // * (x^2-16)*(x-4)*(x+4)

13*(x+4)*(x-4)^2 = 0

( x+4 )

x+4 = 0 // - 4

x = -4

( x-4 )

x-4 = 0 // + 4

x = 4

x in { -4}

x in { 4}

x belongs to the empty set

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