Question

Допоможіть будь ласка. Завдання і варіанти відповідей додаю.

Answer 1

$\displaystyle\bf\\1)\\\\C_{x-1} ^{2} =55\\\\\\\frac{(x-1)!}{2!\cdot(x-1-2)!} =55\\\\\\\frac{(x-1)!}{2!\cdot(x-3)!} =55\\\\\\\frac{(x-3)!\cdot(x-2)\cdot(x-1)}{1\cdot 2\cdot(x-3)!} =55\\\\\\\frac{(x-2)(x-1)}{2} =55\\\\\\x^{2} -3x+2=110\\\\\\x^{2} -3x-108=0\\\\\\x_{1} =12 \ \ \ x_{2} =-9 < 0-neyd\\\\\\Otvet: \ 12$

$\displaystyle\bf\\2)\\\\C_{x+2} ^{2} =45\\\\\\\frac{(x+2)!}{2!\cdot(x+2-2)!} =45\\\\\\\frac{(x+2)!}{2!\cdot x!} =45\\\\\\\frac{x!\cdot(x+1)\cdot(x+2)}{1\cdot 2\cdot x!} =45\\\\\\\frac{(x+1)(x+2)}{2} =45\\\\\\x^{2} +3x+2=90\\\\\\x^{2} +3x-88=0\\\\\\x_{1} =8 \ \ \ x_{2} =-11 < 0-neyd\\\\\\Otvet: \ 8$

$\displaystyle\bf\\3)\\\\A_{x+2} ^{2} =110\\\\\\\frac{(x+2)!}{(x+2-2)!} =110\\\\\\\frac{(x+2)!}{x!} =110\\\\\\\frac{x!\cdot(x+1)\cdot(x+2)}{x!} =110\\\\\\(x+1)(x+2)=110\\\\\\x^{2} +3x+2=110\\\\\\x^{2} +3x-108=0\\\\\\x_{1} =9 \ \ \ ; \ \ \ x_{2} =-12 < 0-neyd\\\\\\Otvet: \ 9$