`a, 2/[3x - 1] = 5/[x+2] ( x ne 1/3 ; x ne - 2)`

`=> 5(3x-1) = 2(x+2)`

`<=> 15x - 5 = 2x + 4`

`<=> 13x = 9`

`<=> x = 9/13 (N)`

Vậy `S={9/13}`

`b, 1/[2x+3] - 3/[x-2] = 5/[(2x+  3)(x-2)] ( x ne -3/2 ; 2)`

`=> x - 2 - 3(2x+3) = 5`

`<=> x - 2 - 6x - 9 = 5`

`<=> -5x = 5 + 2 + 9`

`<=> -5x = 16`

`<=> x= -16/5`

Vậy `S={-16/5}`

`c, [x+3]/[x+1] - [x-1]/x = 3/[x(x+1)] ( x ne 0 ; - 1)`

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

`<=> x^2 + 3x - x^2 + 1 = 3`

`<=> 3x = 2`

`<=> x = 2/3 (N)`

Vậy `S={2/3}`

`d, 7/[x+2] + 2/[x+3] = 1/[x^2 +5x+  6] ( x ne -2;-3)`

`=> 7(x+3) + 2(x+2) = 1`

`<=> 7x+  21 + 2x + 4 = 1`

`<=> 9x = -24`

`<=> x = -9/3 (N)`

Vậy `S={-9/3}`