Đáp án:

 

Giải thích các bước giải:

 `[x+5]/3 -[x-3]/5 =5/[x-3]-3/[x+5](đk:x\ne -5;3)`

`<=>[5(x+5)-3(x-3)]/[15]=[5(x+5)-3(x-3)]/[(x-3)(x+5)]`

`<=>15=(x-3)(x+5)`

`<=>x^2 +2x-15-15=0`

`<=>x^2 +2x-30=0`

`<=>x^2 +2x+1=31`

`<=>(x+1)^2 =31`

`<=>x+1 = ±\sqrt{31}`

`<=>x in { -\sqrt{31}-1; \sqrt{31}-1}`

Vậy `S={-\sqrt{31}-1; \sqrt{31}-1}`

Đáp án+Giải thích các bước giải:

`(x+5)/3-(x-3)/5=5/(x-3)-3/(x+5) (x \ne 3;-5)`

`<=> [5(x+5)]/15-[3(x-3)]/15=[5(x+5)]/[(x-3)(x+5)]-[3(x-3)]/[(x-3)(x+5)]`

`<=> [5(x+5)-3(x-3)]/15=[5(x+5)-3(x-3)]/[(x-3)(x+5)]`

`<=> [5(x+5)-3(x-3)]/15-[5(x+5)-3(x-3)]/[(x-3)(x+5)]=0`

`<=> [5(x+5)-3(x-3)] . (1/15-1/[(x-3)(x+5)])=0`

`TH1:`

`5(x+5)-3(x-3)=0`

`<=> 5x+25-3x+9=0`

`<=> 2x+34=0`

`<=> 2x=-34`

`<=> x=-17 (tm)`

`TH2: 1/15-1/[(x+3)(x+5)]=0`

`<=> [(x-3)(x+5)]/[15(x-3)(x+5)]-15/[15(x-3)(x+5)]=0`

`<=> (x-3)(x+5)-15=0`

`<=> x^2-3x+5x-15-15=0`

`<=> x^2+2x-30=0`

`<=> (x^2+2x+1)-31=0`

`<=> (x+1)^2=31`

`<=> [(x+1=\sqrt{31}),(x+1=-\sqrt{31}):}`

`<=> [(x=\sqrt{31}-1),(x=-\sqrt{31}-1):} (tm)`

Vậy `x \in {+-\sqrt{31}-1; -17}`