1 + 1 + 1 + 1 + 1 + 1 = 6

tuy hơi xấu nhưng đúng

Ta có: A = \(n^2+4n+5=\left(n-1\right)\left(n+1\right)+2\left(2n+3\right)\)

Vì n lẻ suy ra n-1 và n+1 là hai số chẵn liên tiếp suy ra chia hết cho 8 mà 2n+3 lẻ không chia hết cho 4 suy ra\(2\left(2n+3\right)\)không chia hết cho 8⇒ \(n^2+4n+5⋮̸8\)

Ta có: Rtd = 12=R1+R2⇔2R2+R2=12⇔3R2=12⇒R2=4⇒R1=8

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