kết quả bằng 1

\(\frac{2,4+4+2}{8,4}=8,4:8,4=1\)

chuyển \(\frac{2}{20}=\frac{1}{10}\) rồi đến 3/30,4/40,... rồi cứ thế tiếp nhưng ngoại trừ 1/10

\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)

\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)

\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)

\(=1+\frac{28}{5}\)

\(=\frac{33}{5}\)

Ta có:

a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)

b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)

\(=1+\frac{31}{128}=1\frac{31}{128}\)

\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)

\(=\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\right)\cdot5\cdot\frac{1}{5}\)

\(=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)\cdot\frac{1}{5}\)

\(=\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...-\frac{1}{256}\right)\cdot\frac{1}{5}\)

\(=\left(1+1-\frac{1}{256}\right)\cdot\frac{1}{5}\)

\(=\left(2-\frac{1}{256}\right)\cdot\frac{1}{5}\)

\(=\frac{511}{256}\cdot\frac{1}{5}\)

\(=\frac{511}{1280}\)

ĐÚNG NÈ

 \(\frac{10}{2000}:\frac{10}{2000}.\frac{4}{5}+\frac{1}{5}+20\%+80\%\)

=\(1.\frac{4}{5}+\frac{1}{5}+\frac{1}{5}+\frac{4}{5}\)

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

=\(\frac{8}{5}+\frac{2}{5}\)

=\(\frac{10}{5}=2\)

=

Đặt biểu thức trên là A

\(5A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)

Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..+\frac{1}{128}\)

\(2B=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)

\(B=2B-B=1-\frac{1}{128}=\frac{127}{128}\)

\(5A=1+\frac{127}{128}=\frac{255}{128}\Rightarrow A=\frac{5A}{5}=\frac{51}{128}\)

Nguyễn Ngọc Anh Minh đúng rồi