Question

[Bùi Bảo Ngọc]

giúp mình bài này vs ạ. Mình cảm ơn trc

\(\dfrac{x}{x+6}\)+\(\dfrac{3}{x-8}\)=\(\dfrac{-12x+33}{\left(x+6\right)\left(x-8\right)}\)

Answer 1

[Minh Hiếu]

\(\dfrac{x\left(x-8\right)+3\left(x+6\right)}{\left(x+6\right)\left(x-8\right)}=\dfrac{-12x+33}{\left(x+6\right)\left(x-8\right)}\left(đk:x\ne-6;8\right)\)

\(x^2-8x+3x+18=-12x+33\)

\(x^2-5x+18+12x-33=0\)

\(x^2+7x+15=0\)

\(\text{∆}=7^2-4.15=-11< 0\)

⇒ pt vô nghiệm

Answer 2

[Nguyễn Huy Tú]

đk : x khác -6 ; 8 

\(x^2-8x+3x+18=-12x+33\Leftrightarrow x^2+7x-25=0\)

\(\Leftrightarrow x=\dfrac{-7\pm\sqrt{149}}{2}\)