Решите уравнения :1) (x^2-5x)(x+3)(x-8)=02)4x/(x^2+x+3) + 5x/(x^2-5x+3)=1,5

1) $(x^2-5x)(x+3)(x-8)=0;x(x-5)(x+3)(x-8)=0;$
$x=0;x=5;x=-3;x=8.$
2) $\frac{4x}{x^2+x+3}+ \frac{5x}{x^2-5x+3}=1,5;$
ОДЗ: $x^2+x+3 \neq 0, D=1-12=-11<0;x^2-5x+3 \neq 0;D=13;x \neq \frac{5б \sqrt{13} }{2};$
$x^{2} +3=t; \frac{4x}{t+x}+ \frac{5x}{t-5x}= \frac{3}{2};8x(t-5x)+10x(t+x)=3(t+x)(t-5x);$
$8xt-40 x^{2} +10xt+10 x^{2} =3t^2+3xt-15xt-15 x^{2} ;$
$3t^2-30xt+15 x^{2} =0;t^2-10xt+5 x^{2} =0;$
$D_1=25 x^{2} -5 x^{2} =20 x^{2} ;t=5б2 \sqrt{5}x;$
$x^{2} +3=5-2 \sqrt{5}x; x^{2} +2 \sqrt{5}x-2=0;D_1=5+2=7;x=- \sqrt{5}б \sqrt{7};$
$x^{2} +3=5+2 \sqrt{5}x; x^{2} -2 \sqrt{5}x-2=0;D_1=5+2=7;x= \sqrt{5}б \sqrt{7};$
Общий ответ $x=б \sqrt{5}б \sqrt{7}.$