Question

найдите область допустимых значений представленных уравнений:

Answer 1

$1a)\left \{ {{x-2\neq0 } \atop {16-x^{2}\neq0}} \right.\\\\\left \{ {{x\neq 2} \atop {x^{2}\neq16}} \right. \\\\\left \{ {{x\neq2 } \atop {x\neq-4;x\neq4}} \right.\\\\Otvet:\boxedx\in(-\infty;-4)\cup(-4,2)\cup(2;4)\cup(4;+\infty)$

1б)

$\left \{ {{x^{2}-12x+11\neq0} \atop {x^{2}+6x+8\neq0}} \right.\\\\\left \{ {{(x-1)(x-11)\neq0 } \atop {(x+2)(x+4)\neq0}} \right.\\\\\left \{ {{x\neq1;x\neq11} \atop {x\neq-2;x\neq-4}} \right. \\\\Otvet:\boxed{x\in(-\infty;-4)\cup(-4;-2)\cup(-2;1)\cup(1;11)\cup(11;+\infty)}$

$2a)\frac{x^{2}-6 }{x-3}=\frac{x}{x-3}\\\\\frac{x^{2}-6}{x-3}-\frac{x}{x-3}=0\\\\\frac{x^{2}-x-6 }{x-3}=0\\\\\left \{ {{x^{2}-x-6=0 } \atop {x-3\neq0 }} \right.\\\\\left \{ {{x_{1}=3;x_{2}=-2 } \atop {x\neq3}} \right.\\\\Otvet:\boxed{-2}$

2б)

$\frac{x-4}{x}=\frac{2x+10}{x+4}\\\\\frac{x-4}{x}-\frac{2x+10}{x+4}=0\\\\\frac{x^{2}-16-2x^{2}-10x}{x(x+4)}=0\\\\\frac{x^{2}+10x+16 }{x(x+4)}=0\\\\\left \{ {{x^{2}+10x+16=0 } \atop {x\neq0;x+4\neq0}} \right.\\\\\left \{ {{x_{1}=-2;x_{2}=-8} \atop {x\neq0;x\neq-4}} \right.\\\\Otvet:\boxed{-2;-8}$