Question

Прошу помочь пожалуйста 6, 7 и 8...
Даю 80 баллов ​

Answer 1

$6)\ \ \dfrac{20}{x^2-4}-\dfrac{x-3}{x+2}=2\ \ ,\ \ \ ODZ:\ x\ne \pm 2\\\\\\\dfrac{20}{(x-2)(x+2)}-\dfrac{x-3}{x+2}-2=0\ \ ,\ \ \ \dfrac{20-(x-3)(x-2)-2(x^2-4)}{(x-2)(x+2)}=0\\\\\\\dfrac{20-(x^2-5x+6)-2x^2+8}{(x-2)(x+2)}=0\ \ ,\ \ \ \dfrac{-3x^2+5x+22}{(x-2)(x+2)}=0\ \ ,\\\\\\3x^2-5x-22=0\ \ ,\ \ D=289\ ,\ \ x_1=-2\ ,\ x_2=\dfrac{11}{3}\\\\Otvet:\ \ x=\dfrac{11}{3}=3\dfrac{2}{3}\ .$

$7)\ \ \dfrac{3}{x+5}+1=\dfrac{10}{x^2+10x+25}\ \ ,\ \ \ ODZ:\ x\ne -5\ ,\\\\\\\dfrac{3+x+5}{x+5}=\dfrac{10}{(x+5)^2}\ \ ,\ \ \ \dfrac{(x+8)(x+5)-10}{(x+5)^2}=0\ \ ,\\\\\\\dfrac{x^2+13x+30}{(x+5)^2}=0\ \ \to \ \ \ x^2+13x+30=0\ \ ,\\\\\\D=49\ ,\ x_1=-10\ ,\ x_2=-3\\\\Otvet:\ \ x_1=-10\ ,\ x_2=-3$

$8)\ \ \dfrac{2}{x-1}+\dfrac{5}{x-2}=\dfrac{13}{x^2-3x+2}\ \ ,\ \ \ ODZ:\ x\ne 1\ ,\ x\ne 2\ \ ,\\\\\\\dfrac{2}{x-1}+\dfrac{5}{x-2}-\dfrac{13}{(x-1)(x-2)}=0\ \ ,\ \ \ \dfrac{2(x-2)+5(x-1)-13}{(x-1)(x-2)}=0\ ,\\\\\\\dfrac{2x-4+5x-5-13}{(x-1)(x-2)}=0\ \ ,\ \ \ \dfrac{7x-22}{(x-1)(x-2)}=0\ \ ,\ \ \ 7x-22=0\ \ ,\\\\\\x=\dfrac{22}{7}\ \ ,\ \ x=3\dfrac{1}{7}\\\\Otvet:\ \ x=2\dfrac{1}{7}\ .$