0; x=13" alt="\frac{1}{P_(x-4)}=\frac{110}{P_{x-2}};\\ \frac{1}{(x-4)!}=\frac{110}{(x-2)!};\\ (x-2)!=110(x-4)!;\\ (x-4)!(x-3)(x-2)=110(x-4)!;\\ (x-3)(x-2)=110;\\ x^2-5x+6-110=0;\\ x^2-5x-104=0;\\ D=5^2-4*(-104)=441=21^2;\\ x_1=\frac{-(-5)-21}{2*1}=-8;\\ x_2=\frac{-(-5)+21}{2*1}=13;\\ x>0; x=13" align="absmiddle" class="latex-formula">
ответ: 13