Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+....+\frac{1}{10\cdot10}\)
Ta có :
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
.....................................
\(\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)
Ta có :
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{1}{1}-\frac{1}{10}\)
\(\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{9}{10}\)
\(\Rightarrow\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{10\cdot10}< \frac{9}{10}< 1\)
Đặt \(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow B< 1-\frac{1}{10}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
A=20^10+1/20^10-1
A=20^10-1+2/20^10-1
A=20^10-1/20^10-1+2/20^10-1
A=1+2/20^10-1
B=20^10-1/20^10-3
B=20^10-3+2/20^10-3
B=20^10-3/20^10-3+2/20^10-3
B=1+2/20^10-3
Vì 20^10-1>20^10-3 nên 2/20^10-1<2/20^10-3
=>A<B
Ta có: \(20^{10}-1>20^{10}-3\)
\(\Rightarrow\frac{20^{10}-1}{20^{10}-3}>1\)
\(\Rightarrow\frac{20^{10}-1}{20^{10}-3}>\frac{20^{10}-1+2}{20^{10}-3+2}=\frac{20^{10}+1}{20^{10}-1}=B\)
Vậy \(A>B\)
Ta thấy B=20^10-1/20^10-3 là phân số lớn hơn 1.
Theo tính chất nếu a/b>1 thì a/b > a+n/b+n ( n khác 0 )
Ta có : 20^10-1/20^10-3 > 20^10-1+2/20^10-3+2
<=> B > 20^10+1/20^10-3 = A
<=> B > A
Vậy B > A
Lời giải:
$A=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}$
$B=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}$
Vì $20^{10}-1> 20^{10}-3$
$\Rightarrow \frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}$
$\Rightarrow 1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}$
$\Rightarrow A< B$