Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2B-B=1-\frac{1}{64}\)
\(B=\frac{63}{64}\)
A=1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7
=1-1/2+1/2-1/3+1/3-1/4+1/4
=1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
cauB tương tự bạn tự làm nhé
CHÚC BẠN HỌC TỐT
\(B=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdot\cdot\cdot+\frac{6}{97\cdot99}\)
\(\Rightarrow B=3\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{97\cdot99}\right)\)
\(\Rightarrow B=3\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow B=3\cdot\left(1-\frac{1}{99}\right)\)
\(\Rightarrow B=3\cdot\frac{98}{99}\)
\(\Rightarrow B=\frac{98}{33}\)
\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{42}\)
\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{6\cdot7}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{6}-\frac{1}{7}\)
\(\Rightarrow A=1-\frac{1}{7}\)
\(\Rightarrow A=\frac{6}{7}\)
A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 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 + 1/8 - 1/9
= 1 - 1/9 = 8/9
Câu B, C dấu * là nhân hay công vậy?
C=1/2+1/4+1/8+1/16+1/32+1/64+1/128
C=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128
C=1-1/128
C=127/128
D=1/2+1/6+1/12+1/20+1/30+1/42
D=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
D=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
D=1/1-1/7
D=6/7
a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)
\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)
\(=1+1\)
\(=2\)
b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)
\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)
\(=3+2+2\)
\(=7\)
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}\)
\(=\dfrac{6}{7}\)
1) 1/1.2 + 1/2.3 + ... + 1/6.7
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7
= 1 - 1/7
= 6/7
2) 1/2 + 1/6 + 1/12 + .. + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9
= 1 - 1/9
= 8/9
3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)
= \(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)
= \(\frac{1.2....2019}{2.3...2020}\)
= \(\frac{1}{2020}\)
4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)
= \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)
=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)
A = \(\frac{1}{2}-\frac{1}{2^9}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
a) Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32
A = 1/2 + 1/22 + 1/23 + 1/24 + 1/25
2A = 2(1/2 + 22 + 1/23 + 1/24 + 1/25)
2A = 1 + 1/2 + 1/22 + 1/23 + 1/24
2A - A = (1 + 1/2 + 1/22 + 1/23 + 1/24) - (1/2 + 1/22 + 1/23 + 1/24 + 1/25)
A = 1 - 1/25
A = 31/32
b) 2/1.2 + 2/2.3 + 2/3.4 + ... + 2/18 . 19 + 2/19.20
= 2(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/18.19 + 1/19.20)
= 2.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/18 - 1/19 + 1/19 - 1/20)
= 2. (1 - 1/20)
= 2.19/20
= 19/10
\(a,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)
\(\frac{1}{32}=\frac{1}{16}-\frac{1}{32}\)
Thay vào ta có :
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
\(c,\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
...................
\(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
Thay vào ta có :
\(=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)
\(=1-\frac{1}{256}\)
\(=\frac{255}{256}\)