`a) (x+5)/(x^2 - 5x) + (5-x)/(2x^2 + 10x) = (x-5)/(2x^2 - 50) (ĐKXĐ : x \ne 0 ; x \ne +-5)`

`<=> (x+5)/(x . (x-5)) + (5-x)/(2x . (x+5)) = (x-5)/(2 . (x^2 - 25))`

`<=> (2. (x+5).(x+5))/ (2x.(x-5).(x+5)) + ((x-5).(5-x))/(2x . (x+5).(x-5)) = (x. (x-5))/(2x.(x-5).(x+5))`

 `=> 2 . (x+5).(x+5) + (x-5).(5-x) = x . (x-5)`

`<=> (2x + 10) . (x+5) + (5x - x^2 - 25 + 5x) = x^2 - 5x`

`<=> 2x^2 + 10x + 10x + 50 +  5x - x^2 - 25 + 5x = x^2 - 5x`

`<=> x^2 + 30x + 25 = x^2 - 5x`

`<=> x^2 + 30x + 25 - x^2 + 5x = 0`

`<=> 35x + 25 = 0`

`<=> 35x = -25`

`<=> x = -5/7 (TMĐKXĐ)`

Vậy phương trình đã cho có nghiệm `x=-5/7`

`b) (x+1)/(x^2+ x + 1) + (x+4)/(x^2 - x + 1) = (-3)/(x . (x^4 + x^2 + 1))`

`<=> (x . (x+1) . (x^2 - x + 1))/(x. (x^4 + x^2 + 1)) + (x . (x+4).(x^2 + x+1))/(x.(x^4 + x^2+1)) = (-3)/(x . (x^4 + x^2 + 1))`

`=> x . (x+1) . (x^2 - x+1) + x . (x+4) . (x^2 + x +1) = -3 `

`<=> (x^2 + x).(x^2 - x + 1) + (x^2 + 4x).(x^2 + x + 1) = -3`

`<=> (x^4 - x^3 + x^2 + x^3 - x^2 + x) + (x^4 + x^3 + x^2 + 4x^3 + 4x^2 + 4x) = -3`

`<=> (x^4 +x) + (x^4 + 5x^3 + 5x^2 + 4x) = -3`

`<=> x^4 + x + x^4 + 5x^3 + 5x^2 + 4x = -3`

`<=> 2x^4 + 5x^3 + 5x^2 + 5x = -3`

`<=> 2x^4 + 5x^3 + 5x^2 + 5x + 3  =0`

`<=> 2x^4 + 2x^3 + 3x^3 + 3x^2 + 2x^2 + 2x + 3x + 3 = 0`

`<=> 2x^3 . (x + 1) + 3x^2 . (x+1) + 2x . (x+1) + 3 . (x+1) =0`

`<=> (x+1) . (2x^3 + 3x^2 + 2x + 3)  =0`

`<=> (x+1) . [ 2x^2 . (x + 3/2) + 2 . (x+3/2) ] =0`

`<=> (x+1) . (x+3/2) . (2x^2 + 2) = 0`

`<=> (x+1) . (x+3/2) . 2 . (x^2 + 1) = 0`

`<=> (x+1) . (x+3/2) . (x^2 + 1) = 0`

`<=> x + 1 = 0` hoặc `x+3/2 =0` hoặc `x^2 + 1=0`

`+) x + 1=0 <=>x=-1`

`+)x+3/2=0<=>x=-3/2`

`+)x^2 + 1 =0`

`\forall x ` ta có:

`x^2  \ge 0`

`=> x^2 + 1 \ge 0`

`=> x^2 + 1 \ne 0`

Vậy phương trình đã cho có tập nghiệm `S = { -1 ; -3/2 }`