đại số

B

baochauhn1999

$\frac{2x-3}{x+1}-\frac{x^2-5x+22}{x^2-2x-3}=\frac{3x-5}{x-3}$
ĐKXĐ: $x\ne -1;3$
$<=>\frac{(2x-3)(x-3)}{(x-3)(x+1)}-\frac{x^2-5x+22}{(x+1)(x-3)}-\frac{(3x-5)(x+1)}{(x-3)(x+1)}=0$
$<=>\frac{2x^2-9x+9}{(x-3)(x+1)}-\frac{x^2-5x+22}{(x+1)(x-3)}-\frac{3x^2-2x+5}{(x-3)(x+1)}=0$
$<=>-2x^2-2x-18=0$
$<=>x^2+x+9=0$
$<=>(x+\frac{1}{2})^2+\frac{35}{4}=0$ (Vô lí)
$=>PT$ vô nghiệm
 
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