`a)`

`B = (x)/(x+3) + (2x)/(x-3) - (9-3x^2)/(9-x^2)`

(ĐKXĐ: `x \ne \pm 3`)

`= (x)/(x+3) + (2x)/(x-3) - (9-3x^2)/(9-x^2)`

`= (x)/(x+3) + (2x)/(x-3) + (9-3x^2)/(x^2-9)`

`= (x(x-3) + 2x(x+3) + 9 - 3x^2)/((x+3)(x-3))`

`= (x^2 - 3x + 2x^2 + 6x + 9 - 3x^2)/((x+3)(x-3))`

`= (3x + 9)/((x+3)(x-3))`

`= (3(x+3))/((x+3)(x-3))`

`= (3)/(x-3)`

Vậy `B = (3)/(x-3)` với `x \ne \pm 3`

`b)`

`B = (3)/(x-3)` với `x \ne \pm 3`

Ta có: `|2x+1| = 5`

`\text{TH1:}` `2x + 1 = 5`

`2x = 4`

`x = 2 \text{ (TMĐK)}`

`\text{TH2:}` `2x + 1 = -5`

`2x = -6`

`x = -3 \text{ (KTMĐK)}`

Thay `x = 2` vào `B` ta được:

`B = (3)/(2-3)`

`= (3)/(-1)`

`= -3`

Vậy `B = -3` khi `x` thoả mãn `|2x+1| = 5`

`c)`

`B = (3)/(x-3)` với `x \ne \pm 3`

Để `B . (x+10) - 6 = 0` thì

`(3)/(x-3) . (x+10) - 6 = 0`

`(3(x+10))/(x-3) = 6`

`=> 3(x+10) = 6(x-3)`

`3x + 30 = 6x - 18`

`3x = 48`

`x = 16 \text{ (TMĐK)}`

Vậy `x = 16` thì `B . (x+10) - 6 = 0` 

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