DKXD: \(x\ne\pm3\)
\(B=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x}{x+3}+\dfrac{5}{x-3}\right):\left(\dfrac{2x+10}{x+3}-1\right)\)
\(=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x\left(x-3\right)}{x^2-9}+\dfrac{5\left(x+3\right)}{x^2-9}\right):\dfrac{2x+10-x-3}{x+3}\)
\(=\dfrac{x^2+1-x^2+3x+5x+15}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+7}{x+3}\)
\(=\dfrac{\left(8x+16\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+7\right)}\)
\(=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}\)
\(B>0\Rightarrow\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}8x+16>0\\\left(x-3\right)\left(x+7\right)>0\end{matrix}\right.\\\left\{{}\begin{matrix}8x+16< 0\\\left(x-3\right)\left(x+7\right)< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x>-2\\\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 3\\x< -7\end{matrix}\right.\\\left\{{}\begin{matrix}x>3\\x>-7\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}x< -2\\\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 3\\x>-7\end{matrix}\right.\\\left\{{}\begin{matrix}x>3\\x,-7\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
Bài 4 :
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne-7\\x\ne\pm3\end{matrix}\right.\)
Ta có : \(B=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}\)
( Chi tiết xem bạn phía trên :vvv)
- Đặt \(f\left(x\right)=\dfrac{8x+16}{\left(x-3\right)\left(x+7\right)}\)
- Cho \(f\left(x\right)=0\Rightarrow x=-2\)
- Cho ( x - 3 ) ( x + 7 ) = 0
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
- Lập bảng xét dấu :
- Từ bảng xét dấu : Để B = f(x) > 0
\(\Rightarrow x=\left(3;+\infty\right)\)
Vậy...
