Ta có: \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)

\(=\left(1\frac{4}{5}+3\frac{4}{5}\right)+\left(2\frac{5}{7}+4\frac{5}{7}\right)\)

\(=\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)

\(=\frac{28}{5}+\frac{52}{7}=13\frac{1}{35}\)

= ( \(1\frac{4}{5}\)+ \(3\frac{4}{5}\)) + (  \(2\frac{5}{7}\)+  \(4\frac{5}{7}\))         

=        \(4\frac{4}{5}\)           +       \(6\frac{5}{7}\)

=           \(\frac{24}{5}\)         +        \(\frac{47}{7}\)

=  ...... ( tính nốt nhé )

=(1+3)×4/5+(2+4)×5/7

=4×4/5+6×5/7

=16/5+30/7

=16×6/7

gọi biểu thức là A

A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10

=>2A=1+1/2+1/2^2+...+1/2^9

=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10

Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048

A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11

2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10

2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)

A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11

A= 1- 1/2^11

A= 2047/ 2048

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...+\frac{1}{1024}+\frac{1}{2048}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A-A=1-\frac{1}{2048}\)
\(A=\frac{2047}{2048}\)

Bài 1:

\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)

\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)

Bài 2:

\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)

<=>  \(\frac{7}{8}-x=\frac{27}{40}\)

<=>  \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)

Vậy...

bài 2 mình tính sai, sửa

.......

<=>  \(\frac{7}{8}-x=\frac{37}{40}\)

<=>  \(x=\frac{7}{8}-\frac{37}{40}=\frac{-1}{20}\)

Vậy....

\(\frac{4}{5}-\frac{-1}{8}=\frac{7}{8}-x\)

\(\Leftrightarrow x=\frac{7}{8}-\frac{4}{5}-\frac{1}{8}\)

\(\Leftrightarrow x=\frac{-1}{20}\)

giúp mình với nhé

mk cũng lười lắm ,dài quá,mkcũng k hỉu sory

Gọi psố cần tìm là \(\frac{3}{a}\)

-> \(\frac{-1}{2}< \frac{3}{a}< \frac{1}{2}\)

-> \(\frac{-3}{6}< \frac{3}{a}< \frac{3}{6}\)

-> ​\(\frac{+3}{-6}< \frac{3}{a}< \frac{3}{6}\)​

-> a \(\varepsilon\) { -5;-4;-3;-2;-1;0;1;2;3;4;5}

nhớ cho mình nhé . Chúc bạn học tốt 

A = \(\frac{1}{3}-\frac{3}{4}-\frac{-3}{5}+\frac{1}{73}-\frac{1}{36}+\frac{1}{15}+\frac{-2}{9}\)

A = \(\left(\frac{1}{3}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{73}\)

A = \(\left(\frac{3-2}{9}\right)-\left(\frac{27+1}{36}\right)+\left(\frac{9+1}{15}\right)+\frac{1}{73}\)

A  = \(\frac{1}{9}-\frac{7}{9}+\frac{6}{9}+\frac{1}{73}\)

A = \(0+\frac{1}{73}=\frac{1}{73}\)