b: A=1/3+1/9+...+1/3^10

=>3A=1+1/3+...+1/3^9

=>A*2=1-1/3^10=(3^10-1)/3^10

=>A=(3^10-1)/(2*3^10)

c: C=3/2+3/8+3/32+3/128+3/512

=>4C=6+3/2+...+3/128

=>3C=6-3/512

=>C=1023/512

d: A=1/2+...+1/256

=>2A=1+1/2+...+1/128

=>A=1-1/256=255/256

1:

a: =23/27-11/17+4/27+28/17

=23/27+4/27+28/17-11/17

=1+1=2

b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)

=2/3-2/9

=6/9-2/9

=4/9

c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)

=11/5(7/3-1/3)

=11/5*2

=22/5

d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)

e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)

thiếu rồi bạn ơi

C>D thì phải

hoặc C<D

chọn 1 trong2

Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)

A = \(-1+1+\frac{1}{2}\)

A = \(\frac{1}{2}\)

B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)

B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)

B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)

B = \(1+1-\frac{11}{27}\)

B = \(\frac{43}{27}\)

Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)

=> A < B

                       Giải

\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)

\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)

\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)

\(\Leftrightarrow A=-1+1+\frac{1}{2}\)

\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)

\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)

\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)

\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)

\(\Leftrightarrow B=1+\frac{-11}{27}+1\)

\(\Leftrightarrow B=2+\frac{-11}{27}\)

\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)

Từ (1) và (2) suy ra A < B

http://olm.vn/hoi-dap/question/157302.html
\(\text{Đ}\text{ặt}\)\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

             \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

              \(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{49}+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

                \(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)\)

                  \(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\)

so sánh à bạn

ko tính kết quả ý