Прошу, помогите решить хотя бы одно задание, пожалуйста..

Решите задачу:

$1)\; \frac{x^2}{x^2-9}=\frac{12-x}{x^2-9}\; ,\; \; x^2-9\ne 0\; \to \; x\ne \pm 3\\\\x^2=12-x\\\\x^2+x-12=0\; \; \to \; \; x_1=3,\; x_2=-4\\\\Otvet:\; x=40.\\\\2)\; \frac{8}{x-4}+\frac{24}{x}=5\; ,\; \; \left \{ {{x-4\ne 0} \atop {x\ne 0}} \right. \; ,\; \left \{ {{x\ne 4} \atop {x\ne 0}} \right. \\\\\frac{8x+24(x-4)-5x(x-4)}{x(x-4)}=0\\\\\frac{-5x^2+52x-96}{x(x-4)}=0,\; 5x^2-52x+96=0,\; x_1=8\; ,\; x_2=2,4.\\\\3)\; (w-5)^2 \geq w(w-10)\\\\w^2-10w+25-w^2+10w \geq 0\\\\25 \geq 0\; \; verno.\\\\\\c^2+8\geq 2(4c-4)\\\\c^2+8-8c+8\geq 0\\\\c^2-8c+16\geq 0\\\\(c-4)^2\geq 0\; verno$