Ba số tự nhiên liên tiếp có dạng: n; n+1; n + 2 (n \(\in\) N)

Ta cần chứng minh: n(n +1)(n+2) ⋮ 3

nếu n ⋮ 3 ⇒ n(n +1).(n +2) ⋮ 3 (đpcm)

Nếu n = 3k + 1 ⇒ n + 2 =  3k + 1 + 2 = 3k + 3 ⋮ 3

⇒n(n+1).(n+2) ⋮ 3 (đpcm)

Nếu n = 3k + 2 ⇒ n + 1 = 3k + 2 + 1 = 3k + 3 ⋮ 3 

⇒ n.(n + 1).(n +2) ⋮ 3  (đpcm)

A = (\(\dfrac{1}{2}\) + 1).(\(\dfrac{1}{3}\) + 1).(\(\dfrac{1}{4}\) + 1)...(\(\dfrac{1}{99}\) + 1)

A = \(\dfrac{1+2}{2}\).\(\dfrac{1+3}{3}\).\(\dfrac{1+4}{4}\)...\(\dfrac{1+99}{99}\)

A = \(\dfrac{3}{2}\).\(\dfrac{4}{3}\).\(\dfrac{5}{4}\)....\(\dfrac{100}{99}\)

A = \(\dfrac{100}{2}\) \(\times\) \(\dfrac{3.4.5...99}{3.4.5...99}\)

A = 50

\(-136\cdot(64-95)-64\cdot(-95-136)\\=-136\cdot64+(-136)\cdot(-95)+(-64)\cdot(-95)+(-64)\cdot(-136)\\=-136\cdot64+136\cdot95+64\cdot95+64\cdot136\\=(-136\cdot64+64\cdot136)+(136\cdot95+64\cdot95)\\=0+95\cdot(136+64)\\=95\cdot200\\=19000\)

ko dc

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\((-8)\cdot(12+14)+12\cdot(8-14)\\=-8\cdot12+(-8)\cdot14+12\cdot8+12\cdot(-14)\\=(-8\cdot12+12\cdot8)+8\cdot(-14)+12\cdot(-14)\\=-14\cdot(8+12)\\=-14\cdot20\\=-280\)

\(\left(-8\right).\left(12+14\right)+12.\left(8-14\right)\\ =-8.12-8.14+12.8-12.14\\ =\left(12.8-8.12\right)-14.\left(8+12\right)\\ =0-14.20=-280\)

Đề bài thiếu dữ liệu.

câu này hình như bị thiếu