Question

[Thị Thúy Kiều]

Các bạn ơi giúp mình được không ạ? Mình cảm ơn!

Answer 1

[ひまわり(In my personal page there....]

DKXD: \(x\ne\pm3\)

\(B=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x}{x+3}+\dfrac{5}{x-3}\right):\left(\dfrac{2x+10}{x+3}-1\right)\)

  \(=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x\left(x-3\right)}{x^2-9}+\dfrac{5\left(x+3\right)}{x^2-9}\right):\dfrac{2x+10-x-3}{x+3}\)

  \(=\dfrac{x^2+1-x^2+3x+5x+15}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+7}{x+3}\)

  \(=\dfrac{\left(8x+16\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+7\right)}\)

  \(=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}\)

 \(B>0\Rightarrow\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}>0\)

   \(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}8x+16>0\\\left(x-3\right)\left(x+7\right)>0\end{matrix}\right.\\\left\{{}\begin{matrix}8x+16< 0\\\left(x-3\right)\left(x+7\right)< 0\end{matrix}\right.\end{matrix}\right.\)

  \(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x>-2\\\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 3\\x< -7\end{matrix}\right.\\\left\{{}\begin{matrix}x>3\\x>-7\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}x< -2\\\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 3\\x>-7\end{matrix}\right.\\\left\{{}\begin{matrix}x>3\\x,-7\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

Answer 2

[Nguyễn Ngọc Lộc]

phần trên mk không biết chứ phần xét dấu là sai ngoặc hết r nên không tổng hợp lại được đó :vvvv