Question

\(\frac{x+5}{x^2-5x}\)-\(\frac{x+25}{2x^2-50}\)=\(\frac{x-5}{2x^2+10x}\)

Mn giúp em câu này nha.

Answer 1

https://i.imgur.com/TqKLfP9.jpg

Answer 2

\(\frac{x+5}{x^2-5x}-\frac{x+25}{2x^2-50}=\frac{x-5}{2x^2+10x}\) (α)

ĐKXĐ : \(x\ne0;x\ne\pm5\)

Với đk trên, ta có :

(α) ⇔ \(\frac{2\left(x+5\right)^2}{2x\left(x^2-25\right)}-\frac{x^2+25x}{2x\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2x\left(x^2-25\right)}\)

⇔ \(2\left(x^2+10x+25\right)-x^2-25x=x^2-10x+25\)

⇔ \(2x^2+20x+50-x^2-25x=x^2-10x+25\)

⇔ \(2x^2-x^2-x^2+20x-25x+10x=25-50\)

⇔ \(5x=-25\)

⇔ \(x=-5\) (loại)

Vậy : \(S=\varnothing\)