Cho S=\(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{22.27}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{22}-\frac{1}{27}\)

\(=\frac{1}{2}-\frac{1}{27}=\frac{25}{54}\)

kb nha mn!

\(\frac{5}{2}-\frac{5}{7}+\frac{5}{7}-\frac{5}{12}+\frac{5}{12}-\frac{5}{17}+\frac{5}{17}-\frac{5}{22}+\frac{5}{22}-\frac{5}{29}=\frac{5}{2}-0-0-0-0-\frac{5}{29}=\frac{5}{2}-\frac{5}{29}=\frac{145}{58}-\frac{10}{58}=\frac{135}{58}\)

TA CÓ:   \(\frac{1}{n}-\frac{1}{n+5}=\frac{n+5-n}{n\left(n+5\right)}=\frac{5}{n\left(n+5\right)}\)

Thay vào biểu thức trên , ta được:

\(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{907.1002}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{907}-\frac{1}{1002}\)

\(=\frac{1}{2}-\frac{1}{1002}=\frac{250}{501}\)

Đặt A=đã cho

=>\(\frac{5}{3}A=\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+...+\frac{5}{2012\cdot2017}\)

=>\(\frac{5}{3}A=\frac{1}{2}-\frac{1}{2017}\)

Đến đây dễ rồi tự lm tiếp nhé
 

\(=\frac{3}{5}.\left(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{2012.2017}\right)\)

\(=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{2012}-\frac{1}{2017}\right)\)

\(=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{2017}\right)\)

\(=\frac{3}{5}.\left(\frac{2015}{4034}\right)\)

\(=\frac{1209}{4034}\)

Bài này bạn phải học lý thuyết mới làm được nhé!! Chúc bạn zui~^^

Ta có : 5¹ + 5² +...+ 5²⁰¹⁶ 

= (5+5²+5³)+(5⁴+5⁵+5⁶)+...+(5²⁰¹⁴+5²⁰¹⁵+5²⁰¹⁶)

= 5(1+5+5²)+5³(1+5+5²)+...+5²⁰¹³(1+5+5²)

= 5. 31 + 5³.31 +...+5²⁰¹³.31

= 31(5+5²+...+5²⁰¹³) chia hết cho 31

Vây 5¹ + 5² + ...+ 5²⁰¹⁶ chia hết cho 31

Đặt
\(A=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}+\frac{5}{27.32}+\frac{5}{32.37}\)

\(A=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}+\frac{1}{27}-\frac{1}{32}+\frac{1}{32}-\frac{1}{37}\)

\(A=\frac{1}{2}-\frac{1}{37}=\frac{35}{74}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{32}-\frac{1}{37}\)

\(=\frac{1}{2}-\frac{1}{37}\)

\(=\frac{37}{74}-\frac{2}{74}=\frac{35}{74}\)

\(N=2015+\frac{10}{2.7}+\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}\)

     \(=2\left(1007,5+\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}\right)\)

     \(=2\left(1007,5+\frac{7-2}{2.7}+\frac{12-7}{7.12}+\frac{17-12}{12.17}+\frac{22-17}{17.22}\right)\)

     \(=2\left(1007,5+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}\right)\)

     \(=2\left(1007,5+\frac{1}{2}-\frac{1}{22}\right)\)

     \(=2015+1-\frac{1}{11}\)

     \(=\frac{22175}{11}\)

N = \(2015+\frac{10}{2,7}+\frac{10}{7,12}+\frac{10}{12,17}+\frac{10}{17,22}=2021.510611\)

a, \(\frac{1}{2}.B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

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

         \(\frac{1}{2}.B=1-\frac{1}{101}=\frac{100}{101}\)

                  \(B=\frac{100}{101}.2=\frac{200}{101}\)

b, \(\frac{4}{5}.C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)

      \(\frac{4}{5}.C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

          \(\frac{4}{5}.C=1-\frac{1}{105}=\frac{104}{105}\)

                 \(C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)

\(B=\frac{2}{2}\cdot\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+....+\frac{4}{99\cdot101}\right)\)

\(=\frac{4}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)

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

\(=2\cdot\left(1-\frac{1}{101}\right)\)

\(=2\cdot\frac{100}{101}\)

\(=1\frac{99}{101}\)

Bài 2: Mỗi xe ô tô có 4 bánh xe . Hỏi 5 xe ô tô như thế có bao nhiêu bánh xe ?

                                                     Bài giải 

                                          5 xe ô tô như thế có số bánh xe là :

                                            4 x 5= 20 (bánh xe )

                                              Đáp số : 20 bánh xe 

\(B=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{51}\right)=\)\(\frac{250}{51}\)

\(B=5\left(\frac{1}{1}-\frac{1}{51}\right)=\frac{250}{51}\)

\(B=5\left(\frac{51}{51}-\frac{1}{51}\right)=\frac{250}{51}\)

\(B=5.\frac{50}{51}=\frac{250}{51}\)

\(B=\frac{250}{51}=\frac{250}{51}\)