Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
Ta có : \(B=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}\)
=> \(B<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}\)
\(=\frac{7}{8}\)<1
Vậy B < 1
Có 1/2^2 < 1/1.2
1/3^2 <1/2.3
...
1/8^2 < 1<7.8
...tự làm như các phép bình thường
Ta thấy :
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
..............
\(\frac{1}{8^2}< \frac{1}{7\cdot8}\)
=> B
\(=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=1-\frac{1}{8}\)
\(< 1\)
Vậy
Ta có:
1/2^2 < 1/1.2
1/3^2 < 1/2.3
1/4^2< 1/3.4
........................
1/8^2<1/7.8
Vậy B < 1/1.2+1/2.3+1/3.4+....+1/7.8
B< 1-1/8
B<7.8<1
=> B<1
\(B=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+\left(5^6+5^7+5^8\right)\)
\(B=31.1+5^3.31+5^6.31=31.\left(1+5^3+5^6\right)\)
Vậy B chia hết cho 31
=>B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
=>B<1-1/8=7/8<1
Vậy B<1
k cho mk nha