im mồm đi lalisa manoba

 3 bạn nhé ! (@-@)

..., 16, 32, 64, 4096.

1; 1; 2; 2; 4; 16; 32; 64; 4096

lêu lêu đồ hay đi hỏi

đồ nỏ mần được lêu lêu

1234 3246 1535 3456 là đáp án

\(\frac{4}{9}:\frac{5}{7}=\frac{4}{9}\times\frac{7}{5}=\frac{4\times7}{9\times5}=\frac{28}{45}\)

\(\frac{5}{7}:\frac{4}{9}=\frac{5}{7}\times\frac{9}{4}=\frac{5\times9}{7\times4}=\frac{45}{28}\)

\(\frac{1}{3}:\frac{1}{4}=\frac{1}{3}\times4=\frac{4}{3}\)

\(\frac{1}{4}:\frac{1}{3}=\frac{1}{4}\times3=\frac{3}{4}\)

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)

\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)

\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}\times\frac{98}{303}\)

\(A=\frac{49}{303}\)

A= \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ ... + \(\frac{1}{9999}\)

A= \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\) + ... + \(\frac{1}{99.101}\)

2. A= \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) + ... + \(\frac{2}{99.101}\)

2.A = \(\frac{1}{3}\) - \(\frac{1}{5}\)+ \(\frac{1}{5}\)-\(\frac{1}{7}\) + ... + \(\frac{1}{99}\) - \(\frac{1}{101}\)

2.A= \(\frac{1}{3}\) - \(\frac{1}{101}\)

2.A= \(\frac{101}{303}\) - \(\frac{3}{303}\)

2.A= \(\frac{98}{303}\)

A  = \(\frac{98}{303}\) : 2

A  = \(\frac{49}{303}\)

Vay A=\(\frac{49}{303}\)

Lời giải:

$A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{20-19}{19.20}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}$

$=1-\frac{1}{20}=\frac{19}{20}$

19/20

Đặt : \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{128}+\frac{1}{256}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}\)

\(\Rightarrow2A-A=1-\frac{1}{256}\)

\(\Rightarrow A=\frac{255}{256}\)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\cdot\cdot\cdot\frac{99}{100}\)

\(=\frac{1.2....99}{2.3....100}=\frac{1}{100}\)

A=(1-1/2)(1-1/3)(1-1/4)....(1-1/100)

A=1/2.2/3.3/4.....99/100

A=(1.2.3....99)/(2.3.4.....100)

A=1/100

Ta có \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{998\times999}+\frac{1}{999\times1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}\)

\(=\frac{999}{1000}\)