Question

Решите пожалуйста, 18 баллов полное решение!. ​

Answer 1

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

$\left \{ {{x^{2}-8x+7=0 } \atop {x\neq5;x\neq-1}} \right.\\\\x^{2}-8x+7=0\\\\x_{1}=1\\\\x_{2} =7-teorema.Vieta\\\\Otvet:\boxed{1;7}$

$2)\frac{17}{(x-3)(x+4)}-\frac{1}{x-3}-\frac{x}{x+4}=0\\\\\frac{17-(x+4)-x(x-3)}{(x-3)(x+4)}=0\\\\\frac{17-x-4-x^{2}+3x }{(x-3)(x+4)}=0\\\\\frac{-x^{2}+2x+13 }{(x-3)(x+4)}=0\\\\\left \{ {{x^{2}-2x-13=0 } \atop {x\neq3;x\neq-4}} \right.$

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