Question

Помогите пожалуйста по алгебре

Answer 1

$1)\frac{x^{2}}{x-5}+\frac{4x}{5-x}=\frac{5}{x-5}\\\\\frac{x^{2}}{x-5}-\frac{4x}{x-5}-\frac{5}{x-5}=0\\\\\frac{x^{2}-4x-5}{x-5} =0\\\\\left \{ {{x^{2}-4x-5=0 } \atop {x-5\neq0 }} \right.\\\\\\\left \{ {{(x-5)(x+1)=0} \atop {x\neq5 }} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x-5=0\\x+1=0\end{array}\right } \atop {x\neq5 }} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x=5-neyd\\x=-1\end{array}\right} \atop {x\neq5 }} \right.\\\\Otvet:\boxed{-1}$

$2)\frac{4}{x-9}+\frac{9}{x-4}=2\\\\\frac{4(x-4)+9(x-9)}{(x-9)(x-4)}=2\\\\\frac{4x-16+9x-81}{(x-9)(x-4)}=2\\\\\frac{13x-97}{(x-9)(x-4)}=2\\\\2x^{2}-26x+72-13x+97=0;x\neq4;x\neq 9\\\\2x^{2}-39x+169=0\\\\D=(-39)^{2}-4*2*169=1521-1352=169=13^{2}\\\\x_{1} =\frac{39-13}{4}=6,5\\\\x_{2}=\frac{39+13}{4}=13\\\\Otvet:\boxed{6,5;13}$