\(A^3_n+5A^2_n=2\left(n+15\right)\)
ĐK: n ≥ 3 (n∈N)
<=> \(\dfrac{n!}{\left(n-3\right)!}+\dfrac{5.n!}{\left(n-2\right)!}=2\left(n+15\right)\)
<=> \(\dfrac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{\left(n-3\right)!}+\dfrac{5n\left(n-1\right)\left(n-2\right)!}{\left(n-2\right)!}=2\left(n+15\right)\)
<=> \(n\left(n-1\right)\left(n-2\right)+5\left(n-1\right)n-2n-30=0\)
<=> \(n^3+2n^2-5n-30=0\) <=> n=3






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