Bài 104167

$24(A^3_{x+1}-C^{x-4}_x)=23A^4_x \Leftrightarrow 24 \left ( \frac{(x+1)!}{(x-2)!}-\frac{x!}{(x-4)!4!} \right )=23 \frac{x!}{(x-4)!}$ $\Leftrightarrow x^2-6x+5=0 \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = 5\end{array} \right. \Leftrightarrow x=5$

$C^2_n.C^{n-2}_n+2.C^2_n.C^3_n+C^3_n.C^{3-n}_n=100 \Leftrightarrow C^2_n.C^2_n.2C^2_n.C^3_n+C^3_n.C^3_n=100$

Đặt $\begin{cases}x=C^2_n \\ y=C^3_n \end{cases}$, ta có:$x^2+2xy+y^2=100 \Leftrightarrow (x+y)^2=100 \Leftrightarrow x+y=10$

$\Leftrightarrow C^2_n+C^3_n=10 \Leftrightarrow \frac{n!}{2!(n-2)!}+\frac{n!}{3!(n-3)!}=10 \Leftrightarrow n^3-n-60=0$
$\Leftrightarrow (n-4)(n^2+4n+15)=0 \Leftrightarrow n=4$

c. Điều kiện:$x\in N, x\geq3$

$C^1_x+6C^2_x+6C^3_x=9x^2-14x \Leftrightarrow \frac{x!}{1!(x-1)!}+6\frac{x!}{2!(n-2)!}+6\frac{x!}{3!(x-3)!} =9x^2-14x$
$\Leftrightarrow x(x^2-9x+14)=0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\ x= 2\\ x=7\end{array} \right. \Rightarrow x=7$

d. Điều kiện:$A^{y-1}_x:A^y_{x-1}:(C^y_{x-1}+C^{y-1}{x-2})=21:60:10$

$2P_n+6A^2_n-P_nA^2_n=12 \Leftrightarrow (A^2_n-2)(P_n-6)=0 \Leftrightarrow \left[ \begin{array}{l}A^2_n = 2\\ P_n= 6\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\frac{n!}{(n-2)!} = 2\\ n!= 6\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}n^2-n-2 = 0\\ n!= 3!\end{array} \right. \left[ \begin{array}{l}n = -1\\ n= 2 \\ n=3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}n=2\\ n=3\end{array} \right.$