S= 1/2x5 + 1/5x8 + 1/8x11 + 1/11x14 + .... + 1/97x100

S = 1/2 x  1/5 + 1/5 x 1/8 + 1/8 x 1/11 + 1/11 x 1/14 + .......+ 1/97 x 1/100

S =  1/2 x ( 1/5 + 1/5 x 1/8 + 1/8 x 1/11 + 1/11 x 1/14 + .......+ 1/97 ) x 1/100

 S = 1/2 x 1/100

 S = 1/200

~ Hok T ~

\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{97\cdot100}\)

\(S=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+....+\frac{3}{97\cdot100}\right)\)

\(S=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)

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

\(S=\frac{1}{3}\cdot\frac{49}{100}\)

\(S=\frac{49}{300}\)

\(3\times\left(\frac{1}{5\times8}+\frac{1}{8\times11}+....+\frac{1}{97\times100}+x\right)=\frac{319}{100}\)

\(\Rightarrow\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{97\times100}\right)+3\times x=\frac{319}{100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}+3\times x=\frac{319}{100}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{100}+3\times x=\frac{319}{100}\)

\(\Rightarrow\frac{19}{100}+3\times x=\frac{319}{100}\)

\(\Rightarrow3\times x=\frac{319}{100}-\frac{19}{100}\)

\(\Rightarrow3\times x=3\)

\(\Rightarrow x=3:3\)

\(\Rightarrow x=1\)

Vậy x = 1

Dễ, nhưng đúng ra bài này phai là bài của lớp 5 hay lớp 6 chứ bạn

                               Bài làm

Ta có A = 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/14 - 1/17)

          A = 1/3 . ( 1/2 - 1/17 )

          A = 1/3 . 15/34

          A = 5/34

* Có công thức luôn nhé bạn 

Các phân số như vậy có tử số bằng 1 và khoảng cách của hai số ở dưới mẫu bằng nhau  

Một lưu ý nhỏ nữa : các số ở hai mẫu phải có hai số gióng nhau và nằm cạnh nhau ví dụ như: 1/3.5 + 1/5.7 + ....

 => Ta cứ tách ra thành hai phân số như của mình rồi nhân cho phân số có tử số bằng 1 va mẫu số là khoảng cách cua hai số dưới mẫu ban đầu

 chúc bạn luôn làm các dạng bài toán như thế này nè!

mình ko hiểu lắm, 1/3 ở đâu ra vậy

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

\(A=\frac{1}{3}\cdot\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\cdot\frac{24}{49}\)

\(A=\frac{8}{49}\)

Đề sai rồi bạn nhé, đề là như thế này:

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\frac{24}{49}\)

\(A=\frac{24}{147}=\frac{8}{49}\)

A = \(\dfrac{1}{2}\) x 5 + \(\dfrac{1}{5}\) x 8 + \(\dfrac{1}{8}\) x 11 + \(\dfrac{1}{14}\) x 17

A = \(\dfrac{5}{2}\) + \(\dfrac{8}{5}\) + \(\dfrac{11}{8}\) + \(\dfrac{17}{14}\)

A = \(\dfrac{700}{280}\) + \(\dfrac{448}{280}\) + \(\dfrac{385}{280}\) + \(\dfrac{340}{280}\)

\(\Rightarrow\) A = \(\dfrac{1873}{280}\)

A \(=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}\)

A \(=\)\(\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}\right)\)

A \(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\right)\)

A \(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{17}\right)\)

A \(=\dfrac{1}{3}.\dfrac{15}{34}\)

A \(=\dfrac{5}{34}\)

Ta có : \(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\)

\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

\(\Rightarrow S=\frac{3}{7}.\frac{1}{3}=\frac{1}{7}\)

3S= 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/11 - 1/14

3S= 1/2 - 1/14

S= 3/7 / 3

S= 1/7
 

= 2969/440

\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}\)

\(=\frac{1}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}+\frac{3}{17\times20}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\frac{9}{20}\)

\(=\frac{3}{20}\)

_Chúc bạn học tốt_

Đặt \(A=\frac{1}{2x5}+\frac{1}{5x8}+..+\frac{1}{17x20}\)

\(3xA=3x\left(\frac{1}{2x5}+\frac{1}{5x8}+...+\frac{1}{17x20}\right)\)

\(3xA=\frac{3}{2x5}+\frac{3}{5x8}+....+\frac{3}{17x20}\)

\(3xA=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{17}-\frac{1}{20}\)

\(3xA=\frac{1}{2}-\frac{1}{20}\)

\(3xA=\frac{9}{20}\)

\(\Rightarrow A=\frac{3}{20}\)

Đặt \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\)

\(A=\dfrac{3}{2}-\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{14}\)

\(A=\dfrac{3}{2}-\dfrac{3}{14}\)

\(A=\dfrac{21}{14}-\dfrac{3}{14}\)

\(A=\dfrac{18}{14}\)

\(A=\dfrac{9}{7}\)

\(A=1\dfrac{2}{7}\)

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\\ =\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\\ =\dfrac{1}{2}-\dfrac{1}{14}\\ =\dfrac{3}{7}\)

Xin lỗi mình sửa lại

Đặt \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\)

\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\)

\(A=\dfrac{1}{2}-\dfrac{1}{14}\)

\(A=\dfrac{7}{14}-\dfrac{1}{14}\)

\(A=\dfrac{6}{14}=\dfrac{3}{7}\)

Ta có: \(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)

\(\Leftrightarrow x\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)

\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)

\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)

\(\Leftrightarrow x\cdot\frac{3}{7}=\frac{1}{21}\)

\(\Leftrightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}\cdot\frac{7}{3}=\frac{7}{63}=\frac{1}{9}\)

Vậy: \(x=\frac{1}{9}\)