Question

3/(x-3)+3/(x^2-7x+12)=1-x/(x-4)

Answer 1

3/(x-3)+3/(x^2-7x+12)=1-x/(x-4)

3/(x-3)+3/(x-4)(x-3)-(1-x)/(x-4)=0
$frac{3(x-4)+3-(1-x)(x-3)}{(x-4)(x-3)}=0$
$frac{3x-12+3-x+x^2+3-3x}{(x-4)(x-3)}=0$
$frac{x^2-x-6}{(x-4)(x-3)}=0$
D=1+24=25
$frac{(x-3)(x+2)}{(x-4)(x-3)}=0$
$frac{x+2}{x-4}=0$
$x=-2; x neq 4$

Answer 2

спасибо огромное