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11 tháng 5 2022

\(B=\dfrac{49}{2\cdot9}+\dfrac{49}{9\cdot16}+\dfrac{49}{16\cdot23}+...+\dfrac{49}{65\cdot72}\)

\(B=\dfrac{49}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)

\(B=7\cdot\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(B=7\cdot\left(\dfrac{36}{72}-\dfrac{1}{72}\right)\)

\(B=7\cdot\dfrac{35}{72}\)

\(B=\dfrac{\left(7\cdot35\right)}{72}\)

\(B=\dfrac{245}{72}\)

11 tháng 5 2022

\(\dfrac{B}{7}=\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+\dfrac{7}{16\cdot23}+...+\dfrac{49}{65\cdot72}\\ \dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\\ \dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{72}\\ \dfrac{B}{7}=\dfrac{35}{72}\\ B=\dfrac{35}{72}\times7\\ B=\dfrac{245}{72} \)

10 tháng 11 2017

Đề là ntn:

\(A=49\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7.\dfrac{35}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=\dfrac{245}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=\dfrac{735}{72}-\dfrac{7}{36}=\dfrac{735}{72}-\dfrac{14}{36}=\dfrac{721}{36}\)

10 tháng 11 2017

Đề là Thực hiện biến đổi toán học và kết hợp với máy tính . Tính số nghịch đảo của biểu thức ?

10 tháng 11 2017

em mới học lớp 7 thôi ạ

3 tháng 9 2017

\(F=\dfrac{49}{2.9}+\dfrac{49}{9.16}+............+\dfrac{49}{65.72}\)

\(\Leftrightarrow F=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+............+\dfrac{7^2}{65.72}\)

\(\Leftrightarrow F=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+.............+\dfrac{7}{65.72}\right)\)

\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...........+\dfrac{1}{65}-\dfrac{1}{75}\right)\)

\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(\Leftrightarrow F=7.\dfrac{35}{72}=\dfrac{245}{72}\)

\(G=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...........+\dfrac{3}{47.49}\)

\(\Leftrightarrow G=\dfrac{3.2}{1.3.2}+\dfrac{3.2}{3.5.2}+........+\dfrac{3.2}{47.49}\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{47.49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}.\dfrac{48}{49}=\dfrac{72}{49}\)

9 tháng 12 2017

\(\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+\dfrac{7^2}{16.23}+...+\dfrac{7^2}{65.72}\)

\(=7^2\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right)\)

\(=7^2\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)

\(=7^2\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(=49\left(\dfrac{35}{72}\right)\)

\(=\dfrac{1715}{72}\)

9 tháng 12 2017

\(l=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+\dfrac{7^2}{16.23}+...+\dfrac{7^2}{65.72}\)

\(=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+\dfrac{7}{16.23}+...+\dfrac{7}{65.72}\right)\)

\(=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)

\(=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)=7\left(\dfrac{36}{72}-\dfrac{1}{72}\right)=7.\dfrac{35}{72}=\dfrac{245}{72}\)

27 tháng 4 2023

Giúp mình với mình đang cần gấp!!!

 

27 tháng 4 2023

=> D + 49 = (1/49 + 1) + (2/48 + 1) +... (49/1 + 1)

= 50/1 + 50/2 + ... + 50/49

= 50(1/2+1/3+...+1/49) + 50

=> D = 50(1/2 + 1/3 +... + 1/49) + 1

= 50(1/2 + 1/3 +... + 1/49 + 1/50)

=> C/D = 1/50

21 tháng 3 2017

Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)

\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)

18 tháng 5 2022

\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)

\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)

\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)

\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)

11 tháng 3 2017

nhóm từng cái vào r` tính ra

14 tháng 3 2017

Đặt biểu thức cần tính là A, ta có:

A=\(\dfrac{1}{7}\left(\dfrac{7}{3.10}+\dfrac{7}{10.17}+...+\dfrac{7}{73.80}\right)\)

Làm tg tự với những cái khác là ok

11 tháng 5 2017

\(Q=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{47}{3}+\dfrac{48}{2}+\dfrac{49}{1}\\ =\dfrac{1}{49}+1+\dfrac{2}{48}+1+\dfrac{3}{47}+1+...+\dfrac{47}{3}+1+\dfrac{48}{2}+1+1\\ =\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{3}+\dfrac{50}{2}+\dfrac{50}{50}\\ =50\cdot\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+...+\dfrac{1}{3}+\dfrac{1}{2}\right)\\ =50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

\(\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}}{50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

12 tháng 5 2017

bn thiếu dấu ngoặc ở phép thứ 2 rồi