Đáp án:
`a.`Rút gọn:
`A=[\frac{x(3-x)}{(3-x)(3+x)}-1]:[\frac{9-x^2}{x^2+3x-2x-6}+\frac{x-3}{x-2}-\frac{x+2}{x+3}]`
`= [\frac{x}{3+x}-1]:[\frac{9-x^2}{(x+3)(x-2)}+\frac{x-3}{x-2}-\frac{x+2}{x+3}]`
`= \frac{x-3-x}{3+x}:\frac{9-x^2(x+3)(x-3)-(x-2)(x+2)}{(x+3)(x-2)}`
`= \frac{-3}{3+x}:\frac{9-x^2+x^2-9-x^2+4}{(x+3)(x-2)}`
`= \frac{-3}{3+x}:\frac{4-x^2}{(x+3)(x-2)}`
`= \frac{-3}{3+x}:\frac{-(x-2)(2+x)}{(x+3)(x-2)}`
`= \frac{-3}{3+x}:\frac{-(2+x)}{x+3}`
`= -3xx\frac{1}{-2-x}`
`= -3xx\frac{1}{-(2+x)}`
`= \frac{3}{2+x}`
`b.`Ta có: `2A+1=0`
`<=> 2A=-1`
`<=> A=-1/2`
Để `A=-1/2` thì `\frac{3}{2+x}=-1/2`
`<=> 3xx2=-(2+x)`
`<=> 6=-2-x`
`<=> 6+2+x=0`
`<=> 8+x=0`
`<=> x=-8`
`c.`Để `A<0` thì `\frac{3}{2+x}<0` (vì `3>0`)
`<=> 2+x<0`
`<=> x< -2`
