a) \(\left(\frac{3}{x+3}+\frac{1}{x-3}-\frac{18}{9-x^2}\right)\left(x-2\right)\)
= \(\left[\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{1\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{18}{x^2-9}\right]\left(x-2\right)\)
= \(\left[\frac{3x-9}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{18}{\left(x-3\right)\left(x+3\right)}\right]\left(x-2\right)\)
=\(\frac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}.\left(x-2\right)\)
=\(\frac{4x+12}{\left(x-3\right)\left(x+3\right)}.\left(x-2\right)\)
=\(\frac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\left(x-2\right)\)
=\(\frac{4}{x-3}.\frac{x-2}{1}\)
=\(\frac{4\left(x-2\right)}{x-3}\)
Vậy ...
b) Ta có : \(\frac{4\left(x-2\right)}{x-3}=4+\frac{4}{x-3}\) [ ĐKXĐ : x\(\ne\pm3\) ]
Để A \(\in Z\) <=> \(\frac{4}{x-3}\) \(\in Z\)
<=> x - 3 \(\inƯ_4=\left\{\pm1,\pm2,\pm4\right\}\)
Ta có bảng sau :
| x - 3 | -1 | 1 | -2 | 2 | -4 | 4 |
| x | 2 | 4 | 1 | 5 | -1 | 7 |
| Nhận xét : | tm | tm | tm | tm | tm | tm |
Vậy ...
c) Để B<0, B>0 thì
x - 3 \(\ne0\)
và \(x+3\ne0\)
\(\Leftrightarrow x\ne\pm3\)
Vậy ...
