so sanh A=9^10;B=8^9+7^9+.......+2^9+1^9
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 387420490 ; B = 1000000001
vậy B lớn hơn A
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}\)
\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)
\(\frac{-9}{10^{2010}}>\frac{-19}{10^{2010}}\)
\(\Rightarrow\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}>\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1=B\)
a/
\(\frac{2001}{2004}=\frac{2004-3}{2004}=1-\frac{3}{2004}=1-\frac{1}{668}.\)
\(\frac{39}{40}=\frac{40-1}{40}=1-\frac{1}{40}\)
Ta có \(40< 668\Rightarrow\frac{1}{40}>\frac{1}{668}\Rightarrow1-\frac{1}{40}< 1-\frac{1}{668}\Rightarrow\frac{39}{40}< \frac{2001}{2004}\)
b/
\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1=B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
9/10 ..... 9 + 3/10 + 3
9/10....... 12/13
9/10=117/130
12/13 = 120/130
Vì 120/130 > 117/130 Nên 9 + 3/10 + 3 > 9/10
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(1990^10 + 1990^9) và 1991^10
1990^10 + 1990^9 = 1990.1990^9 + 1990^9 = 1991^9 < 1991^10
--> (1990^10 + 1990^9) < 1991^10
mọi người giúp mk với nhé