Chứng tỏ:
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1995}-\dfrac{1}{1996}=\dfrac{1}{996}+\dfrac{1}{997}+...+\dfrac{1}{1990}\)
Giúp Mình vs
Những câu hỏi liên quan
1,CMR:\(1-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{1990}=\dfrac{1}{996}+\dfrac{1}{997}+\dfrac{1}{1990}\)
1,CMR:
B,\(1-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{1990}=\dfrac{1}{996}+\dfrac{1}{997}+\dfrac{1}{990}\)
1,CMR:\(1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{1990}=\dfrac{1}{996}+\dfrac{1}{997}+...+\dfrac{1}{990}\)
cho A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.....+\(\dfrac{1}{2^{2020}}\)+\(\dfrac{1}{2^{2021}}\). Chứng tỏ rằng A<\(\dfrac{1}{2}\)
Giúp vs ạ cần gấp
làm vào bài đừng có dùng ngoặc kép như tui nha,tui làm minh họa cho bạn hiểu
Đúng 0
Bình luận (0)
\(\dfrac{1}{\begin{matrix}1\times&2\end{matrix}}+\dfrac{1}{\begin{matrix}2\times&3\end{matrix}}+\dfrac{1}{\begin{matrix}3\times&4\end{matrix}}+...........+\dfrac{1}{x\times\left(x+1\right)}=\dfrac{996}{997}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{996}{997}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)= \(\dfrac{996}{997}\) \(1-\dfrac{1}{x+1}\) = \(\dfrac{996}{997}\)
\(\dfrac{1}{x+1}\) = \(1-\dfrac{996}{997}\)
\(\dfrac{1}{x+1}\) =\(\dfrac{1}{997}\)
\(\Rightarrow\) x + 1 = 997
x = 997 - 1
x = 996
Vậy x = 996
Đúng 0
Bình luận (0)
\(\dfrac{x+1}{1998}\)+ \(\dfrac{x+2}{1997}\)=\(\dfrac{x+3}{1996}\)+\(\dfrac{x+4}{1995}\)
=>(x+1/1998+1)+(x+2/1997+1)=(x+3/1996+1)+(x+4/1995+1)
=>x+1999=0
=>x=-1999
Đúng 0
Bình luận (0)
Bài 1:Chứng tỏ rằng:Bdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+dfrac{1}{5^2}+dfrac{1}{6^2}+dfrac{1}{7^2}dfrac{1}{8^2}1Bài 2:Chứng tỏ rằng:Edfrac{3}{4}+dfrac{8}{9}+dfrac{15}{16}+...+dfrac{2499}{2500}1Bài 3:Chứng tỏ rằng:1dfrac{2011}{2020^2+1}+dfrac{2021}{2020^2+2}+dfrac{2021}{2020^3+3}+...+dfrac{2021}{2020^3+2020} 2
Đọc tiếp
Bài 1:Chứng tỏ rằng:B=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)\(\dfrac{1}{8^2}\)<1
Bài 2:Chứng tỏ rằng:E=\(\dfrac{3}{4}\)+\(\dfrac{8}{9}\)+\(\dfrac{15}{16}\)+...+\(\dfrac{2499}{2500}\)<1
Bài 3:Chứng tỏ rằng:1<\(\dfrac{2011}{2020^2+1}\)+\(\dfrac{2021}{2020^2+2}\)+\(\dfrac{2021}{2020^3+3}\)+...+\(\dfrac{2021}{2020^3+2020}\)< 2
1:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
Đúng 0
Bình luận (0)
7 tháng 7 2021 lúc 16:51
Bài 3:
c) Chứng tỏ rằng \(\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{43}+\dfrac{1}{44}>\dfrac{5}{6}\)
Giúp mik vs! Thanks nhiều nha!
21 tháng 2 2023 lúc 23:01
Chứng tỏ rằng:
\(\dfrac{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{199.200}}{\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}}=1\)
Giúp mình với📖
