Ta có:

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\)\(\frac{1}{19}\)

\(B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+...+\frac{1}{19}\right)\)

\(\Rightarrow B>\left(\frac{1}{15}+\frac{1}{15}+\frac{1}{15}+...+\frac{1}{15}\right)+\left(\frac{1}{20}+...+\frac{1}{20}\right)\)

     \(B>\frac{4}{5}+\frac{1}{5}\)

    \(B>1\)\(\left(đpcm\right)\)

\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(A=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-A=\left(\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(-A=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+\frac{1}{8}-\frac{1}{7}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)

\(-A=\frac{1}{10}-1=\frac{-9}{10}\Rightarrow A=\frac{9}{10}\)

\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)

Vậy A=-79/90

A=1/1.2-1/2.3-1/3.4-1/4.5-1/5.6-1/6.7-1/7.8-1/8.9-1/9.10(mình làm ngược lại nha)

A=1/1-1/2-1/2-1/3-1/3-1/4-1/4-1/5-1/5-1/6-1/6-1/7-1/7-1/8-1/8-1/9

A=1/1-1/9

A=8/9

\(A=0,4\left(3\right)+0,6\left(2\right)\cdot2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,5\left(8\right)}:\frac{50}{53}\)

\(A=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{3+2}{6}:\frac{53}{90}\cdot\frac{53}{50}\)

\(A=\frac{13}{30}+\frac{14}{9}-\frac{5}{6}\cdot\frac{90}{53}\cdot\frac{53}{50}\)

\(A=\frac{39}{90}+\frac{140}{90}-\frac{2}{3}\)

\(A=\frac{179}{90}-\frac{60}{90}=\frac{119}{90}\)

\(A=1,3\left(2\right)\)

#)Giải :

Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(A=\frac{1}{5}-\frac{1}{10}\)

\(A=\frac{1}{10}\)

cho mk hỏi chút là tại sao ta lấy 1/5 - 1/6 r + 1/6....

\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\)

\(A=\frac{1}{5}-\frac{1}{10}\)

\(A=\frac{2}{10}-\frac{1}{10}=\frac{1}{10}\)

A= 1+ 1/2 + 1/2² + ... + 1/2²⁰¹² 

﴾1/2﴿A= 1/2+1/2²+...+1/2²⁰¹³

A‐﴾1/2﴿A= ﴾1+ 1/2 + 1/2² + ... + 1/2²⁰¹² ﴿ ‐ ﴾ 1/2+1/2²+...+1/2²⁰¹³ ﴿

﴾1/2﴿A = 1 ‐ 1/2²⁰¹³ 

A= ﴾1‐ 1/2²⁰¹³ ﴿ : 1/2

A= 2 ‐ 1/2²⁰¹²

\(A=2-\frac{1}{2^{2012}}\)

A= 1+ 1/2 + 1/2² + ... + 1/2²⁰¹² 

﴾1/2﴿A= 1/2+1/2²+...+1/2²⁰¹³

A‐﴾1/2﴿A= ﴾1+ 1/2 + 1/2² + ... + 1/2²⁰¹² ﴿ ‐ ﴾ 1/2+1/2²+...+1/2²⁰¹³ ﴿

﴾1/2﴿A = 1 ‐ 1/2²⁰¹³ 

A= ﴾1‐ 1/2²⁰¹³ ﴿ : 1/2

A= 2 ‐ 1/2²⁰¹²