a, đk x khác 2 ; 3
\(K=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}=\dfrac{2x-9-\left(x^2-9\right)+\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b, Ta có \(K-1=\dfrac{x+1}{x-3}-1=\dfrac{x+1-x+3}{x-3}=\dfrac{4}{x-3}>0\Rightarrow x>3\)
c, \(\dfrac{x-3+4}{x-3}=1+\dfrac{4}{x-3}\Rightarrow x-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| x-3 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 4 | 2 | 5 | 1 | 7 | -1 |
d, Ta có \(\dfrac{x+1}{x-3}=2\Leftrightarrow x+1=2x-6\Leftrightarrow x=7\)
e, Ta có \(x^2-3x+2=0\Leftrightarrow x=1;x=2\)
Với x = 1 thì K = 2/-2 =-1
Với x = 2 thì K = 3/-1 = -3
