Question

Решите уравнения :
1) (x^2-5x)(x+3)(x-8)=0
2)4x/(x^2+x+3) + 5x/(x^2-5x+3)=1,5

Answer 1

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 $x=б sqrt{5}б sqrt{7}.$