a, Để B xác định
\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\4-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
\(b,B=\dfrac{3}{x-2}+\dfrac{-2}{x+2}-\dfrac{x-14}{4-x^2}\)
\(=\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{-2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3x+6-2x+4+x-14}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
c, Đẻ B có giá trị nguyên
\(\Leftrightarrow2⋮x+2\Leftrightarrow x+2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Ta có bẳng sau:
| \(x+2\) | 1 | -1 | 2 | -2 |
| 2 | -1 | -3 | 0 | -4 |
Vậy \(x\in\left\{-1;-3;0;-4\right\}\) thì B có giá trị nguyên
