Giải thích các bước giải:

`a)`

`(x^2 + 1)/(x + 1) + (x^2 + 2)/(x - 2) = -2 (x`$\neq$ `-1 ; x` $\neq$ `2)`

`=> ((x^2 + 1)(x - 2) + (x^2 + 2)(x + 1))/((x + 1)(x - 2)) = -2`

`=> (x^2 + 1)(x - 2) + (x^2 + 2)(x + 1) = -2(x + 1)(x - 2)`

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

`=> (x^3 + x^3) + (x^2 - 2x^2) + (x + 2x) + (2 - 2) + 2(x^2 - x - 2) = 0`

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

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

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

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

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

Ta có `:`

`2x^2 + 3x + 4 = 2(x^2 + 3/2x + 9/16) + 23/8 = 2(x + 3/4)^2 + 23/8 > 0`

`=> x - 1 = 0`

`=> x = 1` (tm)

Vậy `x = 1`

`b)`

`(x + 1)/(x - 1) = (3x - 2)/(3x + 1)` `(x`$\neq$ `1 ; x` $\neq$ `-1/3)`

`=> (x + 1)(3x + 1) = (3x - 2)(x - 1)`

`=> 3x^2 + 4x + 1 = 3x^2 - 5x + 2`

`=> 4x + 1 = -5x + 2`

`=> 9x = 1`

`=> x = 1/9` (tm)

Vậy `x = 1/9`