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a) A=3+32+33+34+35+36+....+328+329+330A=3+32+33+34+35+36+....+328+329+330
⇔A=(3+32+33)+(34+35+36)+....+(328+329+330)⇔A=(3+32+33)+(34+35+36)+....+(328+329+330)
⇔A=3(1+3+32)+34(1+3+32)+....+328(1+3+32)⇔A=3(1+3+32)+34(1+3+32)+....+328(1+3+32)
⇔A=3.13+34.13+....+328.13⇔A=3.13+34.13+....+328.13
⇔A=13(3+34+....+328)⋮13(dpcm)⇔A=13(3+34+....+328)⋮13(dpcm)
b) A=3+32+33+34+35+36+....+325+326+327+328+329+330A=3+32+33+34+35+36+....+325+326+327+328+329+330
⇔A=(3+32+33+34+35+36)+....+(325+326+327+328+329+330)⇔A=(3+32+33+34+35+36)+....+(325+326+327+328+329+330)
⇔A=3(1+3+32+33+34+35)+....+325(1+3+32+33+34+35)⇔A=3(1+3+32+33+34+35)+....+325(1+3+32+33+34+35)
⇔A=3.364+....+325.364⇔A=3.364+....+325.364
⇔A=364(3+35+310+....+325)⇔A=364(3+35+310+....+325)
⇔A=52.7(3+35+310+....+325)⋮52(dpcm)
Ta có \(\left(n+3\right)^2\ge0\forall x\) \(\Rightarrow4-\left(n+3\right)^2\le4\forall x\)
Dấu "=" xảy ra khi \(n+3=0\Leftrightarrow n=-3\)
Vậy \(A_{min}=4\) khi \(x=-3\)
\(\left|n-3\right|\ge0,\forall n\\ \Rightarrow A=\left|n-3\right|+2\ge2\)
Dấu \("="\Leftrightarrow n=3\)
\(\left|n-3\right|+2\ge2\forall n\)
Dấu '=' xảy ra khi n=3
A
6 thuộc A