Verified answer

[tex]\displaystyle\bf\\1)\\\\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\\\\\\(x+1)\cdot(x+2)=90\\\\x^{2} +3x+2-90=0\\\\x^{2} +3x-88=0\\\\x_{1} =8 \ \ \ ; \ \ \ x_{2} =-11 < 0- \ ne \ podxodit\\\\Otvet \ : \ 8\\\\2)\\\\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[/tex]

[tex]\displaystyle\bf\\(x-2)\cdot(x-1)=110\\\\x^{2} -3x+2-110=0\\\\x^{2} -3x-108=0\\\\x_{1} =12 \ \ \ ; \ \ x_{2} =-9- < 0 \ ne \ podxodit\\\\Otvet \ : \ 12\\\\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)\cdot(x+2)=110\\\\x^{2} +3x+2-110=0\\\\x^{2} +3x-108=0\\\\x_{1} =9 \ \ ; \ \ x_{2} =-12- < 0 \ ne \ podxodit\\\\Otvet \ : \ 9[/tex]