Đáp án+Giải thích các bước giải:

 `a)A=((6)/(x^{2}-9)-(5x)/(3x-x^{2})+(1)/(x+3)):(2x-1)/(x^{2}-3x)`

`( đ k : x\ne0;x\ne\pm3;x\ne 1/2)`

`=[(6)/((x-3).(x+3))+(5x)/(x.(x-3))+(1)/(x+3)].(x^{2}-3x)/(2x-1)`

`=[(6x)/(x.(x-3).(x+3))+(5x.(x+3))/(x.(x-3).(x+3))+(x.(x-3))/(x.(x-3).(x+3))].(x^{2}-3x)/(2x-1)`

`=[(6x+5x^{2}+15x+x^{2}-3x)/(x.(x-3).(x+3))].(x^{2}-3x)/(2x-1)`

`=((6x^{2}+18x).x.(x-3))/(x.(x-3).(x+3).(2x-1))`

`=(6x.(x+3))/((x+3).(2x-1))`

`=(6x)/(2x-1)`

Vậy `A=(6x)/(2x-1)` với `x\ne0;x\ne\pm3;x\ne 1/2`

`d)A=(-2)/(3)`

`(6x)/(2x-1)=(-2)/(3)` `( đ k : x\ne0;x\ne\pm3;x\ne 1/2)`

`6x.3=-2.(2x-1)`

`18x+2.(2x-1)=0`

`18x+4x-2=0`

`22x-2=0`

`22x=2`

`x=1/11(tmđk)`

Vây `x=1/11` để `P=(-2)/(3)`