Question

100 балов Спасибо Помогите

Answer 1

$\displaystyle\bf\\A_{x} ^{5} =72A_{x-2} ^{3} \\\\\\\frac{x!}{(x-5)!} =72\cdot\frac{(x-2)!}{(x-2-3)!} \\\\\\\frac{(x-5)!\cdot(x-4)\cdot(x-3)\cdot(x-2)\cdot(x-1)\cdot x}{(x-5)!} =72\cdot\frac{(x-2)!}{(x-5)!} \\\\\\(x-4)\cdot(x-3)\cdot(x-2)\cdot(x-1)\cdot x=72\cdot\frac{(x-5)!\cdot(x-4)\cdot(x-3)\cdot(x-2)}{(x-5)!} \\\\\\(x-4)\cdot(x-3)\cdot(x-2)\cdot(x-1)\cdot x=72(x-4)\cdot(x-3)\cdot(x-2)\\\\\\(x-1)\cdot x=72\\\\x^{2} -x-72=0\\\\x_{1} =9\\\\x_{2} =-8 < 0- \ ne \ podxodit\\\ Otvet: 9$