`40/(x-6)+40/(x+10)=80/x` (1)
ĐKXĐ: `x\ne 6; -10; 0`
`(1) <=> 1/(x-6)+1/(x+10)=2/x`
`<=> (x+10+x-6)/((x-6)(x+10))=2/x`
`<=> (2x+4)/(x^2+10x-6x-60)=2/x`
`<=> (2x+4)/(x^2+4x-60)=2/x`
`<=> x(2x+4)=2(x^2+4x-60)`
`<=> 2x^2+4x=2x^2+8x-120`
`<=> 2x^2+8x-120-2x^2-4x=0`
`<=> 4x-120=0`
`<=> 4(x-30)=0`
`<=> x-30=0`
`<=> x=30 (TM)`
Vậy `x=30`