Question

Решите пожалуйста, 18 баллов полное решение!. ​

Answer 1

$1)frac{10}{(x-5)(x+1)}+frac{x}{x+1}-frac{3}{x-5}=0\\frac{10+x(x-5)-3(x+1)}{(x-5)(x+1)}=0\\frac{10+x^{2}-5x-3x-3 }{(x-5)(x+1)}=0\\frac{x^{2}-8x+7 }{(x-5)(x+1)}=0$

$left { {{x^{2}-8x+7=0 } atop {xneq5;xneq-1}} right.\\x^{2}-8x+7=0\\x_{1}=1\\x_{2} =7-teorema.Vieta\\Otvet:boxed{1;7}$

$2)frac{17}{(x-3)(x+4)}-frac{1}{x-3}-frac{x}{x+4}=0\\frac{17-(x+4)-x(x-3)}{(x-3)(x+4)}=0\\frac{17-x-4-x^{2}+3x }{(x-3)(x+4)}=0\\frac{-x^{2}+2x+13 }{(x-3)(x+4)}=0\\left { {{x^{2}-2x-13=0 } atop {xneq3;xneq-4}} right.$

$x^{2} -2x-13=0\\D=(-2)^{2}-4*(-13)=4+52=56 =(2sqrt{14})^{2}\\x_{1}=frac{2-2sqrt{14}}{2}=1-sqrt{14}\\x_{2} =frac{2+2sqrt{14}}{2}=1+sqrt{14}\\Otvet:boxed{1-sqrt{14} ;1+sqrt{14}}$