Question

(n-1)!<=72(n-3)! решите неравенство

Answer 1

$(n-1)!leq 72(n-3)!\\star ; ; (n-1)!=(n-3)!, cdot , (n-2)cdot (n-1); ; to \\(n-2)cdot (n-1)leq 72\\n^2-3n-70 leq 0\\n_{1,2}=frac{3pm 17}{2}; ,; ; n_1=-7notin N; ,; ; n_2=10in N\\n^2-3n-70leq 0; ; to ; ; nin [-7,10, ]\\(n-1)in N^circ ; ,; ; (n-3)in N^circ ; ; Rightarrow ; ; (n-3)geq 0; ,; ; ngeq 3\\Otvet:; ; nin [, 3,10, ]; .$