1 + 1 = 2
2 + 2 = 4
k cho tớ nha
thanks
1 + 1 = 2 ; 2 + 2 = 4.
Kích mình nha!
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
>, <, = ?
| 2 + 2 … 5 | 2 + 1 … 1 + 2 | 3 + 1 … 3 + 2 |
| 2 + 3 … 5 | 2 + 2 …1 + 2 | 3 + 1 … 1 + 3 |
| 5 + 0 … 5 | 2 + 0 …1 + 2 | 1 + 4 …. 4 + 1 |
Tính:
| 3 + 1 + 1 = … | 1 + 2 + 2 =… | 2 + 1 + 1 =… |
| 1 + 3 + 1 =… | 2 + 2 +1 =… | 2 + 1 + 2 =… |
+, - ?
| 1… 2 = 3 | 2 … 1 = 3 | 1…1 = 2 | 1…4 = 5 |
| 3…1 = 2 | 3…2 = 1 | 2…1 = 1 | 2…2 = 4 |
1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2x22
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+..+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
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 \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}\)
\(\Rightarrow VT< \frac{1}{2^2}\left(2-\frac{1}{50}\right)=\frac{1}{2}-\frac{1}{2^2.50}< \frac{1}{2}\left(Đpcm\right)\)
Tính:
2 + 1 + 1 = 3 + 1 + 1 = 1 + 2 + 2 =
1 + 2 + 1 = 1 + 3 + 1 = 2 + 2 + 1 =
Tính:
2 - 1 = 3 - 1 = 1 + 1 = 1 + 2 =
3- 1 = 3 - 2 = 2 - 1 = 3 - 2 =
3 -2 = 2 - 1 = 3 - 1 = 3 - 1 =