1+1=2
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ờ thì 1 + 1 = 2 . Chứ chẳng nhẽ 1 + 1 = 3 à ?
1+1+1+1+1+1+1+1+1+1+1+1+11++1+1+1+11+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1-2+2+2+2+2+2+2-12221+21
1+0+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+11+1+1++1+11+1+1+1+1+1+1+1+2+2+2+2+2+2222+33333+444+555+666+777+888+999-{1+0+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+11+1+1++1+11+1+1+1+1+1+1+1+2+2+2+2+2+2222+33333+444+555+666+777+888+999-}
1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2=
5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5=
1-1-1-1-1-1-1-1-1-1-1-1-+2-2-2-2-2-2-2-2+4+4-3-13+12345678
Tính?
| 1 + 1 =… | 1 + 2 =… | 2 + 2 = … | 1 + 1 =… |
| 2 + 1 =… | 1 + 3 =… | 3 + 1 =… | 1 + 2 =… |
| 3 + 1 =… | 1 + 1 =… | 1 + 3 =… | 2 + 1 =… |
Tính:
2 + 1 + 1 = 3 + 1 + 1 = 1 + 2 + 2 =
1 + 2 + 1 = 1 + 3 + 1 = 2 + 2 + 1 =
Tính:
| 3 + 1 + 1 = … | 1 + 2 + 2 =… | 2 + 1 + 1 =… |
| 1 + 3 + 1 =… | 2 + 2 +1 =… | 2 + 1 + 2 =… |
Tính:
1 + 2 = 1 + 1 = 1 + 2 = 1 + 1 + 1 =
1 + 3 = 2 - 1 = 3 - 1 = 3 - 1 - 1 =
1 + 4 = 2 + 1 = 3 - 2 = 3 - 1 + 1 =
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+..+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Có \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\) \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)....v........v............ \(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
Cộng lại \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}\)
\(\Rightarrow VT< \frac{1}{2^2}\left(2-\frac{1}{50}\right)=\frac{1}{2}-\frac{1}{2^2.50}< \frac{1}{2}\left(Đpcm\right)\)
