`̣(x^2 + 5)/(25-x^2 ) = 3/(x+5) + x/(x-5)` Tìm `x`

Đáp án:

`S={1}`

Giải thích các bước giải:

`(x^2+5)/(25-x^2)=3/(x+5)+x/(x-5)(x\ne+-5)`
`<=>(-x^2-5)/(x^2-25)=(3.(x-5))/((x+5)(x-5))+(x.(x+5))/((x+5)(x-5))`
`<=>(-x^2-5)/((x+5)(x-5))-(3x-15)/((x+5)(x-5))-(x^2+5x)/((x+5)(x-5))=0`
`<=>(-x^2-5-3x+15-x^2-5x)/((x+5)(x-5))=0`
`<=>(-2x^2-8x+10)/((x+5)(x-5))=0`
`=>-2x^2-8x+10=0`
`<=>-2.(x^2+4x-5)=0`
`<=>x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>x.(x+5)-(x+5)=0`
`<=>(x-1)(x+5)=0`
`<=>[(x=1),(x=-5(KTM)):}`
Vậy `S={1}`