ĐKXĐ: `x != 0, x != -1, x != -2, x != -3`

`5/x + 4/(x+1) = 3/(x+2) + 2/(x+3)`

`(5/x + 1) + (4/(x+1) + 1) = (3/(x+2) + 1) + (2/(x+3) + 1)`

`(5 + x)/x + (5 + x)/(x + 1) = (5 + x)/(x + 2) + (5 + x)/(x + 3)`

`(5 + x)/x + (5 + x)/(x + 1) - (5 + x)/(x + 2) - (5 + x)/(x + 3) = 0`

`(x + 5) * (1/x + 1/(x + 1) - 1/(x + 2) - 1/(x + 3)) = 0`

`5 + x = 0` hoặc `1/x + 1/(x + 1) - 1/(x + 2) - 1/(x + 3) = 0`

TH1: `x + 5 = 0`

`x = -5` (`text{tmđk}`)

TH2: `1/x - 1/(x + 2) + 1/(x + 1) - 1/(x + 3) = 0`

`2 / (x(x + 2)) + 2 / ((x + 1)(x + 3)) = 0`

`1 / (x^2 + 2x) + 1 / (x^2 + 4x + 3) = 0`

`(x^2 + 4x + 3 + x^2 + 2x) / ((x^2 + 2x)(x^2 + 4x + 3)) = 0`

`(2x^2 + 6x + 3) / ((x^2 + 2x)(x^2 + 4x + 3)) = 0`

`2x^2 + 6x + 3 = 0`

Ta có: `a = 2, b = 6, c = 3`

`Delta = 6^2 - 4 * 2 * 3 = 12 > 0`

Vì `Delta > 0` nên phương trình có `2` nghiệm phân biệt:

`x_1 = (-b + sqrt{Delta}) / (2a) = (-6 + sqrt{12}) / (2 * 2) = (-6 + sqrt{12}) / 4 = (-3 + sqrt{3}) / 2` (`text{tmđk}`)

`x_2 = (-b - sqrt{Delta}) / (2a) = (-6 - sqrt{12}) / (2 * 2) = (-6 - sqrt{12}) / 4 = (-3 - sqrt{3})/2` (`text{tmđk}`)

Vậy `text{S} = {-5;(-3 +- sqrt{3})/2}`