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3 tháng 2 2017

 ta có A =\(\frac{1}{5\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot15}+...+\frac{1}{605\cdot608}\)

3A =\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{605\cdot608}\)

3A =\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\)

3A=\(\frac{1}{5}-\frac{1}{608}\)

3A=\(\frac{603}{3040}\)A =\(\frac{201}{3040}\)

3 tháng 2 2017

Đặt A=\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\)

      3A=\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\right)\)

      3A=\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\right)\)

      3A=3.\(\left(\frac{1}{5}-\frac{1}{608}\right)\)

       A=\(\frac{201}{3040}\)

18 tháng 3 2017

a, \(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)

=> x + 3 = 308

     x = 308 - 3

     x = 305

b, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)

\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}.\frac{3984}{1993}\)

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1992}{1993}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{1992}{1993}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)

=> x + 1 = 1993

     x = 1993 - 1

     x = 1992

18 tháng 3 2017

a ,\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(x=308-3\)

\(x=305\)

30 tháng 1 2017

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)

30 tháng 1 2017

b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)

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

\(=\frac{5}{4}.\frac{4n}{12n+9}\)

\(=\frac{5n}{12n+9}\)

( sai đề )

9 tháng 4 2015

đề sai rồi 13/15.4 chứ

B.1/7=1/7.(5/2.1+4/1.11+3/11.2+1/2.15+13/15.14)

B.1/7=5/2.1.7+4/1.11.7+3/11.2.7+1/2.15.7+13/15.4.7)

B.1/7=5/2.7+4/7.11+3/11.14+1/14.15+13/15.28

B.1/7=1/2-1/7+1/7-1/11+1/11-1/14+1/14-1/15+1/15-1/28

B.1/7=1/2-1/28=13/28

B=13/28:1/7

B=13/4

14 tháng 6 2016

a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)

                                                              \(=1-\frac{1}{32}=\frac{31}{32}\)

b) \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)\

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)

12 tháng 5 2019

Đề là cm S>1 nha bạn!

\(S=\frac{9}{2.5}+\frac{9}{5.8}+...+\frac{9}{29.32}\)

\(=3\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)

\(=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(=3\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(=3.\frac{15}{32}\)

\(=\frac{45}{32}>1\)

\(\Leftrightarrow S>1\)

12 tháng 5 2019

\(S=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{29\cdot32}\)

Cách 1 : Vì hiệu hai thừa số đều là 3 = 5 - 2 = 8 - 5 = ... = 32 - 29 nên phân tích tử 9 = 3 . 3

Ta có : \(S=3\left[\frac{3}{2\cdot5}+\frac{3}{7\cdot9}+...+\frac{3}{29\cdot32}\right]=3\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{32}\right]\)

\(=3\left[\frac{1}{2}-\frac{1}{32}\right]=3\left[\frac{16}{32}-\frac{1}{32}\right]=3\cdot\frac{15}{32}=\frac{45}{32}\)

Mà \(\frac{45}{32}>1\)=> S không thể bé hơn 1

Cách 2 : Nhận xét : \(\frac{9}{2\cdot5}=\frac{3}{2}-\frac{3}{5};\frac{9}{5\cdot8}=\frac{3}{5}-\frac{3}{8};...\)

Vậy ta có : \(S=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{29\cdot32}=\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+...+\frac{3}{29}-\frac{3}{32}\)

\(=\frac{3}{2}-\frac{3}{32}=\frac{3\cdot16}{32}-\frac{3}{32}=\frac{48}{32}-\frac{3}{32}=\frac{45}{32}\)

Tự so sánh , mà S đâu bé hơn 1 ???

7 tháng 2 2017

a)\(\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{x}-\frac{1}{x+3}\right)\)=\(\frac{101}{1540}\)

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)

=\(\frac{101}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}\)=\(\frac{101}{1540}:\frac{1}{3}\)=\(\frac{303}{1540}\)

\(\frac{1}{x+3}\)=\(\frac{1}{5}-\frac{303}{1540}\)=\(\frac{1}{308}\)

\(\Rightarrow\)x+3=308

\(\Rightarrow\)x=308-3=305

b)Mk chưa nghĩ ra

7 tháng 2 2017

b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2}{9}\)

\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{x+1-6}{6\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)

\(\Rightarrow9x-45=6x+6\)

\(\Rightarrow3x=51\)

\(\Rightarrow x=17\)

Vậy x = 17