1) 2) 3) = 42

Решите задачу:

$1)\;C_{x-1}^2=6\\\frac{(x-1)!}{2!(x-1-2)!}=6\\\frac{(x-1)!}{2!(x-3)!}=6\\\frac{(x-2)(x-1)}{2!}=6\\x^2-3x+2=12\\x^2-3x-10=0\\D=9+4\cdot10=49\\x_1=-2,\;x_2=5\\2)\;A_{x+2}^2=6x(x+1)\\\frac{(x+2)!}{(x+2-2)!}=6x(x+1)\\\frac{(x+2)!}{x!}=6x(x+1)\\\frac{(x+1)(x+2)}{x}=6x(x+1)\\\frac{(x+2)}{x}=6x\\x+2=6x^2\\6x^2-x-2=0\\D=1+4\cdot6\cdot2=49\\x_1=-\frac12,\;x_2=\frac23$
$3.\;\frac{A_x^4\cdot P_{x-4}}{P_{x-2}}=42\\\frac{\frac{x!}{(x-4)!}\cdot(x-4)!}{(x-2)^!}=42\\\frac{x}{(x-2)!}=42\\(x-1)x=42\\x^2-x-42=0\\D=1+4\cdot42=169\\x_1=-6,\;x_2=7$