Giúp mình bài 1 với ạ

Bài `1:`

`a) A=(5x+1)/(2x-3) . (x+2)/(25x^2-1) - (8-3x)/(25x^2-1) : (2x-3)/(5x+1)`

ĐKXĐ: `{(2x-3 \ne 0),(25x^2 -1 \ne 0),(5x+1 \ne 0):}`

`<=> {(x \ne 3/2),(5x-1 \ne 0),(5x+1 \ne 0):}`

`<=> {(x \ne 3/2) , (x \ne 1/5),(x \ne -1/5):}`

Vậy `ĐKXĐ` là: `x \ne 3/2` và `x \ne +- 1/5`

`b) A=(5x+1)/(2x-3) . (x+2)/(25x^2-1) - (8-3x)/(25x^2-1) : (2x-3)/(5x+1)`

`A=((5x+1) . (x+2))/((2x-3) . (5x-1)(5x+1)) - ((8-3x)(5x+1))/((5x-1)(5x+1) . (2x-3))`

`A=(x+2)/((2x-3) . (5x-1)) - (8-3x)/((5x-1)(2x-3)`

`A=(x+2 - 8+3x)/((5x-1)(2x-3))`

`A=(4x-6)/((5x-1)(2x-3))`

`A=(2(2x-3))/((5x-1)(2x-3))`

`A=(2)/(5x-1)`

Ta có: `A . B = (x+2)/(5x-1)`

`=> (2)/(5x-1) . B = (x+2)/(5x-1)`

`=> B = (x+2)/(5x-1) : 2/(5x-1)`

`=> B = ((x+2) . (5x-1))/((5x-1) . 2)`

`=> B = (x+2)/(2)`

`c)` Tại `x=3/5` thì ta được:

`B=(x+2)/2`

`B=(3/5 +2)/2`

`B = 13/5 : 2`

`B=13/10`

Vậy `B=13/10` tại `x=3/5`