`g) 1/(2x-3)-3/(x(2x-3))=5/x`  `(đk: xne0; xne3/2)`

`⇒ x-3=5(2x-3)`

`⇔ x-3=10x-15`

`⇔ x-10x=-15+3`

`⇔ -9x=-12`

`⇔ x=4/3 (t//m)`

                           Vậy `S={4/3}`

`h) (2x)/(x-1)+4/(x+2x-3)=(2x-5)/(x+3)`  `(đk: xne1; xne-3)`

`⇔ (2x)/(x-1)+4/((x-1)(x+3))=(2x-5)/(x+3)`

`⇒ 2x(x+3)+4=(2x-5)(x-1)`

`⇔ 2x^2+6x+4=2x^2-7x+5`

`⇔ 2x^2+6x-2x^2+7x=5-4`

`⇔ 13x=1`

`⇔ x=1/13 (t//m)`

                                Vậy `S={1/13}`

 

`g, 1/(2x-3) - 3/(x(2x-3)) = 5/x`

ĐKXĐ: `x \ne 0, x \ne 3/2`

` ⇔ x/(x(2x-3)) - 3/(x(2x-3)) = (5(2x-3))/(x(2x-3)`

` ⇔ x/(x(2x-3)) - 3/(x(2x-3)) - (5(2x-3))/(x(2x-3)=0`

` ⇔ (x-3)/(x(2x-3)) - (10x-15)/(x(2x-3)=0`

` ⇔ (x-3-10x+15)/(x(2x-3))=0`

` ⇔ (-9x+12)/(x(2x-3))=0`

` ⇔ -9x+12=0`

` ⇔ -9x=-12`

` ⇔ x=4/3`(t/m)

Vậy `x=4/3`

`h, (2x)/(x-1) + 4/(x^2+2x-3)=(2x-5)/(x+3)`

ĐKXĐ: `x \ne 1, x \ne -3`

`⇔ (2x)/(x-1) + 4/(x^2+2x-3)-(2x-5)/(x+3)=0`

`⇔ (2x(x+3))/((x-1)(x+3)) + 4/((x-1)(x+3))-((2x-5)(x-1))/((x-1)(x+3))=0`

`⇔ (2x^2+6x)/((x-1)(x+3)) + 4/((x-1)(x+3))-(2x^2-7x+5)/((x-1)(x+3))=0`

`⇔ (2x^2+6x+4-2x^2+7x-5)/((x-1)(x+3))=0`

`⇔ (13x-1)/((x-1)(x+3))=0`

`⇔ 13x-1=0`

`⇔ 13x=1`

`⇔ x=1/13`(t/m)

Vậy `x=1/13`