Question

Даю 50 балов !!!!!!!!!

Answer 1

Ответ: $\bf m=8$

Объяснение:

$\displaystyle 12C^{m-1}_{m+3}=55A^2_{m+1} \ \ ; \ \ m\in \mathbb N \\\\\\ 12\cdot \frac{(m+3)!}{(m-1)!(m+3-(m-1))!} =55\cdot \frac{(m+1)!}{(m+1-2)!}\\\\\\ 12\cdot\frac{(m+3)!}{(m-1)!\cdot 4!} =55\cdot \frac{(m+1)!}{(m-1)!} \ \ \ \ \ | \ \cdot (m-1)! \quad ; \ \ \ m\neq 1 \\\\\\ (m+3)!:2=55(m+1)! \\\\\\ (m+3)(m+2)(m+1)!=110(m+1)! \\\\\\ (m+3)(m+2)=110\\\\\\ m^2+5m+6-110=0\\\\\\ m^2+5m-104=0 \\\\\left \{ {{m_1+m_2=-5} \atop {m_1m_2=-104 }} \right. <=>m_1=-13 \ \ ; \ \ \boxed{m_2=8 }$