1. giải phương trình a) 5 4 x-5 ——– + ——- = ———– x – 3 x+3 x^2-9 b) x+2 1

a) ĐK: `x \ne \pm 3`

`5/(x-3)+4/(x+3)=(x-5)/(x^2-9)`

`<=> 5(x+3)+4(x-3)=x-5`

`<=> 5x+15+4x-12=x-5`

`<=> 9x+3=x-5`

`<=>8x=-8`

`<=>x=-1(TM)`

Vậy `S={-1}`

b) ĐK: `x \ne 0 ; x \ne 2`

`(x+2)/(x-2)-1/x=2(x^2-2x)`

`<=> (x+2)/(x-2)-1/x=2/(x(x-2))`

`<=> x(x+2)-(x-2)=2`

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

`<=>x^2+x=0`

`<=>x(x+1)=0`

`<=>` \(\left[ \begin{array}{l}x=0(L)\\x=-1(TM)\end{array} \right.\) 

Vậy `S={-1}`

c) ĐK: `x \ne 1 ; x \ne 3`

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

`<=> 2/(x-1)-1/(x-3)=(3x)/((x-1)(x-3))`

`<=> 2(x-3)-(x-1)=3x`

`<=>2x-6-x+1=3x`

`<=>-2x=6`

`<=>x=-3 (TM)`

Vậy `S={-3}`