Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =(-1)+(-1)+...+(-1)=-1011
b: =(-5)+(-5)+...+(-5)=-175
1:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
=>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
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
Giải:
Dễ thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(.................\)
\(\dfrac{1}{8^2}=\dfrac{1}{8.8}< \dfrac{1}{7.8}\)
Cộng vế theo vế ta được:
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=1-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
Vậy \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{8^2}< 1\) (Đpcm)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=\frac{2-1}{1.2}+......+\frac{8-7}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{8}=1-\frac{1}{8}< 1\)
ta có điều phải chứng minh
a: M=-2021+2021-68-68+17
=-119
b: B=(-1)+(-1)+...+(-1)
=-1x500
=-500
c: C=(1-2-3+4)+(5-6-7+8)+...+(997-998-999+1000)
=0
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2021^2}\)
\(< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2020.2021}\)
\(=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+...+\frac{2021-2020}{2020.2021}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=\frac{1}{4}-\frac{1}{2021}< \frac{1}{4}\)