`a)(x+5)/(x^2-5x)-(x+25)/(2x^2-50)=(x-5)/(2x^2+10x)` $(ĐKXĐ: x≠0;x≠±5)$

$⇔ \dfrac{x+5}{x(x-5)}-\dfrac{x+25}{2(x^2-25)}=\dfrac{x-5}{2x(x+5)}$

$⇔ \dfrac{x+5}{x(x-5)}-\dfrac{x+25}{2(x-5)(x+5)}=\dfrac{x-5}{2x(x+5)}$

$⇔ \dfrac{2(x+5)(x+5)}{2x(x-5)(x+5)}-\dfrac{x(x+25)}{2x(x-5)(x+5)}=\dfrac{(x-5)(x-5)}{2x(x-5)(x+5)}$

$⇔ \dfrac{2(x+5)^2}{2x(x-5)(x+5)}-\dfrac{x^2+25x}{2x(x-5)(x+5)}=\dfrac{(x-5)^2}{2x(x-5)(x+5)}$

$⇔ \dfrac{2(x^2+10x+25)}{2x(x-5)(x+5)}-\dfrac{x^2+25x}{2x(x-5)(x+5)}=\dfrac{x^2-10x+25}{2x(x-5)(x+5)}$

$⇒ 2x^2+20x+50-(x^2+25x)=x^2-10x+25$

$⇔ 2x^2+20x+50-x^2-25x=x^2-10x+25$

$⇔ x^2-5x+50=x^2-10x+25$

$⇔ x^2-x^2-5x+10x+50-25=0$

$⇔ 5x+25=0$

$⇔ 5x=-25$

$⇔ x=-5$ (không thỏa mãn điều kiện xác định)

Vậy phương trình có tập nghiệm `S={∅}`

Đáp án:

`S=∅`

Giải thích các bước giải:

`d)`
`(x+5)/(x^2-5x)-(x+25)/(2x^2-50)=(x-5)/(2x^2+10x)(x\ne0;+-5)`
`<=>(x+5)/(x.(x-5))-(x+25)/(2.(x^2-25))-(x-5)/(2x.(x+5))=0`
`<=>(2.(x+5)^2)/(2x.(x-5)(x+5))-(x.(x+25))/(2x.(x-5)(x+5))-(x-5)^2/(2x.(x+5)(x-5))=0`
`<=>(2.(x+5)^2-x.(x+25)-(x-5)^2)/(2x.(x-5)(x+5))=0`
`<=>(2.(x^2+10x+25)-x^2-25x-x^2+10x-25)/(2x.(x-5)(x+5))=0`
`<=>(2x^2+20x+50-x^2-25x-x^2+10x-25)/(2x.(x-5)(x+5))=0`
`<=>(5x+25)/(2x.(x-5)(x+5))=0`
`=>5x+25=0`
`<=>x=-5(text{KTM})`
Vậy `S=∅`