Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)
\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)
\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)
\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)
1a
75/100+18/21+19/32+1/4+3/21+13/32
= 3/4 +6/7+19/32+1/4+1/7+13/32
= (3/4+1/4)+(19/32+13/32)+(6/7+1/7)
= 1+1+1=3
1b
22/5+51/9+11/4+3/5+1/3+1/4
=22/5+17/3+11/4+3/5+1/3+1/4
=(22/5+3/5)+(17/3+1/3)+(11/4+1/4)
=25/5+18/3+12/4
=5+6+3
=14
75/100+18/21+19/32+1/4+3/21+13/32
= 18/21 + 3/21 + 19/32 + 13/32 + 75/100 + 1/4
= (18/21 + 3/21) + (19/32 + 13/32) + (3/4 + 1/4)
= 21/21 + 32/32 + 4/4
= 1 + 1 + 1
= 3
b, \(K =\) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\dfrac{3}{4}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{18}{21}+\dfrac{3}{21}\right)+\left(\dfrac{19}{32}+\dfrac{13}{32}\right)\)
\(K = \) \(1 + 1 + 1\)
\(K = \) \(3\)
Bài 1:
5; (-23) + 105
= 105 - 23
= 82
6; 78 + (-123)
= 78 - 123
= - (123 - 78)
= - 45
bài1
1)2763 + 152 = 2915
2)-7 +(-14)
=-(14 +7)
=-21