Toán Lớp 11: Giải phương trình sau: $P_n.A^2_n+72=6(A^2_n+2P_n)$

Question

Toán Lớp 11:  Giải phương trình sau: $P_n.A^2_n+72=6(A^2_n+2P_n)$

maianhnhiquynh · Answer

Điều kiện $n ge 2, n in  mathbb N$   $ begin{array}{l} {P_n}.A_n^2 + 72 = 6 left( {A_n^2 + 2{P_n}}  right)     Leftrightarrow {P_n}A_n^2 + 72 - 6A_n^2 - 12{P_n} = 0     Leftrightarrow {P_n}A_n^2 - 6A_n^2 - 12{P_n} + 72 = 0     Leftrightarrow A_n^2 left( {{P_n} - 6}  right) - 12 left( {{P_n} - 6}  right) = 0     Leftrightarrow  left( {{P_n} - 6}  right) left( {A_n^2 - 12}  right) = 0     Leftrightarrow  left[  begin{array}{l} {P_n} = 6   A_n^2 = 12  end{array}  right.  Leftrightarrow  left[  begin{array}{l} n! = 6    dfrac{{n!}}{{ left( {n - 2}  right)!}} = 12  end{array}  right.     Leftrightarrow  left[  begin{array}{l} n = 3   n left( {n - 1}  right) = 12  end{array}  right.     Leftrightarrow  left[  begin{array}{l} n = 3   {n^2} - n - 12 = 0  end{array}  right.     Leftrightarrow  left[  begin{array}{l} n = 3(tm)   n = 4(tm)   n =  - 3(L)  end{array}  right.  Rightarrow  left[  begin{array}{l} n = 3   n = 4  end{array}  right.  end{array}$

haiyemy · Answer

Bạn kham khảo ĐK:n>=2;n inNN Ta có: P_n . A_n^2+72=6(A_n^2+2P_n) <=>n! . (n!)/((n-2)!)+72=6[(n!)/((n-2)!)+2.n!] <=>n!.(n-1).n+72=6[(x-1)n+2.n!] <=>(n!-6)(n^2-n-12)=0 <=>[(n^2-n-12=0),(n!-6=0):} <=>[(x=4(t//m)),(n=-3(L)),(n=3(t//m)):} Vậy n={4;3}