6/(x²-4ч+3)-(13-7x)/(1-x)-3/(x-3)=0
6/(x-1)(x-3)+(13-7x)/(x-1)-3/(x-3)=0
ОЗ (x-1)(x-3)≠0⇒x≠1,x≠3
6+(13-7x)(x-3)-3(x-1)=0
6+13x-39-7x²+21x-3x+3=0
-7x²+31x-30=0
7x²-31x+30=0
D=961-840=121
x1=(31-11)/14=10/7
x2=(31+11)/14=3 не удов усл
Ответ х=10/7
(2/(x+1)+10/(x²-3x-4)+3x/(x-4)):(3x+2)/3=3/(x-4)
1)2/(x+1)+10/[(x+1)9x-4)+3x/9x-4)=(2x-8+10+3x²+3x)/[(x+1)(x-4)]=
=(3x²+5x+2)/[(x+1)(x-4)]=(3x+2)9x+1)/[(x+1)(x-4)]=(3x+2)/(x-4)
2)(3x+2)/(x-4) : (3x+2)/3=(3x+2)/(x-4) * 3/(3x+2)=3/(x-4)