b.
Điều kiện: \(x\ge2\)
\(P_x.A_x^2+72=6\left(A_x^2+2P_x\right)\)
\(\Leftrightarrow P_x.A_x^2-6A_x^2-12P_x+72=0\)
\(\Leftrightarrow A_x^2\left(P_x-6\right)-12\left(P_x-6\right)=0\)
\(\Leftrightarrow\left(P_x-6\right)\left(A_x^2-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}P_x=6\\A_x^2=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
c.
Điều kiện: \(n\ge3\)
\(2\left(A_n^3+3A_n^2\right)=P_{n+1}\)
\(\Leftrightarrow\dfrac{2.n!}{\left(n-3\right)!}+\dfrac{6.n!}{\left(n-2\right)!}=\left(n+1\right)!\)
\(\Leftrightarrow\dfrac{2}{\left(n-3\right)!}+\dfrac{6}{\left(n-2\right)!}=n+1\)
- Kiểm tra với \(n=3\) ko thỏa mãn, \(n=4\) thỏa mãn
- Với \(n\ge5\Rightarrow\left\{{}\begin{matrix}\left(n-3\right)!\ge2!=2\\\left(n-2\right)!\ge3!=6\end{matrix}\right.\)
\(\Rightarrow\dfrac{2}{\left(n-3\right)!}+\dfrac{6}{\left(n-2\right)!}\le\dfrac{2}{2}+\dfrac{6}{6}=2< n+1=6\)
\(\Rightarrow\) Không tồn tại \(n\ge5\) thỏa mãn
Vậy \(n=4\)
d.
ĐK: \(x\ge2\)
\(2A_x^2+50=A_{2x}^2\)
\(\Leftrightarrow\dfrac{2.x!}{\left(x-2\right)!}+50=\dfrac{\left(2x\right)!}{\left(2x-2\right)!}\)
\(\Leftrightarrow2x\left(x-1\right)+50=2x\left(2x-1\right)\)
\(\Leftrightarrow2x^2=50\)
\(\Leftrightarrow x=5\)
