Question

Допоможіть будьласка терміново

Answer 1

Ответ: x=8.

Объяснение:

ОДЗ: х+1≥5    х≥4      х≥3      ⇒      х∈[4;+∞).

$\displaystyle\\C^5_{x+1}=\frac{3A^3_x}{8}\\\\\frac{(x+1)!}{(x+1-5)*5!}=\frac{3*\frac{x!}{(x-3)!} }{8} \\\\\\\frac{(x-4)!*(x-3)*(x-2)*(x-1)*x*(x+1)}{(x-4)!*120} =\frac{3*(x-3)!*(x-2)*(x-1)*x}{8*(x-3)!}\\\\\\\frac{(x-3)*(x-2)*(x-1)*x*(x+1)}{120} =\frac{3*(x-2)*(x-1)*x}{8} \\\\\\\frac{(x-3)*(x+1)}{120} =\frac{3}{8} \ |*120\\\\x^2-2x-3=3*15\\\\x^2-2x-3=45\\\\x^2-2x-48=0\\\\x^2-8x+6x-48=0\\\\x*(x-8)+6*(x-8)=0\\\\(x-8)*(x+6)=0\\\\x-8=0\\\\x_1=8\\\\x+6=0\\\\x_2=-6.$