`A=(8 2/7-4 2/7)-3 4/9`

`=8+2/7-4-2/7-3-4/9`

`=4-3-4/9`

`=1-4/9=5/9`

`B=(10 2/9-6 2/9)+2 3/5`

`=10+2/9-6-2/9+2+3/5`

`=4+2+3/5`

`=6+3/5=33/5`

Bài 2:

`a)5 1/2*3 1/4`

`=11/2*13/4`

`=143/8`

`b)6 1/3:4 2/9`

`=19/3:38/9`

`=19/3*9/38=3/2`

`c)4 3/7*2`

`=31/7*2`

`=62/7`

Bài 1:

\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\) 

\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\) 

\(A=4-\dfrac{31}{9}\) 

\(A=\dfrac{5}{9}\) 

\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\) 

\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\) 

\(B=4+\dfrac{13}{5}\) 

\(B=\dfrac{33}{5}\)

Bài 2:

a) \(5\dfrac{1}{2}.3\dfrac{1}{4}=\dfrac{11}{2}.\dfrac{13}{4}=\dfrac{11.13}{2.4}=\dfrac{143}{8}\) 

b) \(6\dfrac{1}{3}:4\dfrac{2}{9}=\dfrac{19}{3}:\dfrac{38}{9}=\dfrac{19}{3}.\dfrac{9}{38}=\dfrac{3}{2}\) 

c) \(4\dfrac{3}{7}.2=\dfrac{31}{7}.2=\dfrac{31.2}{7}=\dfrac{62}{7}\)

A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

A = 2A - A = \(1-\frac{1}{2^9}\)

Kết quả của phép tính \(\frac{2}{3}+\frac{-1}{6}\)

\(\frac{2}{3}+\frac{-1}{6}\)

\(=\)\(\frac{4}{6}+\frac{-1}{6}\)

\(=\)\(\frac{3}{6}=\frac{1}{2}\)

Vậy chọn đáp án \(a.\frac{1}{2}\)

\(A=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{79}\right)\)

\(\Rightarrow A=\frac{3}{5}.\frac{5}{7}...\frac{77}{79}\)

\(\Rightarrow A=\frac{3}{99}\)

\(\Rightarrow A=\frac{1}{3}\)

P=( 1-1/7) * 9 1- 2/7) * (1-3/7 ) .... ( 1- 10/7)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

\(2A=2+\frac{2}{2}+\frac{2}{2^2}+...+\frac{2}{2^9}\)

\(2A=2+1+\frac{1}{2}+...+\frac{2}{2^8}\)

=>\(A=2+\frac{-1}{29}\)

=>\(A=\frac{57}{29}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

=>  \(2A=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)

=> \(A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

=> \(A=2+1+\frac{1}{2}+...+\frac{1}{2^8}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)

=> \(A=2-\frac{1}{2^9}\)