Tính tổng B = \(\dfrac{49}{2.9}+\dfrac{49}{9.16}+\dfrac{49}{16.23}+...+\dfrac{49}{65.72}\)
HELP
49 x \(\dfrac{1}{2.9}\)+ \(\dfrac{1}{9.16}\)+\(\dfrac{1}{16.23}\)+...............+\(\dfrac{1}{65.72}\):\(\dfrac{1}{3}\) - \(\dfrac{7}{36}\)
Đề là ntn:
\(A=49\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=7.\dfrac{35}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=\dfrac{245}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)
\(A=\dfrac{735}{72}-\dfrac{7}{36}=\dfrac{735}{72}-\dfrac{14}{36}=\dfrac{721}{36}\)
49 x \(\dfrac{1}{2.9}\)+ \(\dfrac{1}{9.16}\)+\(\dfrac{1}{16.23}\)+...............+\(\dfrac{1}{65.72}\):\(\dfrac{1}{3}\) - \(\dfrac{7}{36}\)
F=49/2.9+49/9.16+49/16.23+...+49/65.72
G=3/1.3+3/3.5+3/5.7+..+3/47.49
\(F=\dfrac{49}{2.9}+\dfrac{49}{9.16}+............+\dfrac{49}{65.72}\)
\(\Leftrightarrow F=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+............+\dfrac{7^2}{65.72}\)
\(\Leftrightarrow F=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+.............+\dfrac{7}{65.72}\right)\)
\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...........+\dfrac{1}{65}-\dfrac{1}{75}\right)\)
\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)
\(\Leftrightarrow F=7.\dfrac{35}{72}=\dfrac{245}{72}\)
\(G=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...........+\dfrac{3}{47.49}\)
\(\Leftrightarrow G=\dfrac{3.2}{1.3.2}+\dfrac{3.2}{3.5.2}+........+\dfrac{3.2}{47.49}\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{47.49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{49}\right)\)
\(\Leftrightarrow G=\dfrac{3}{2}.\dfrac{48}{49}=\dfrac{72}{49}\)
Tính: \(\dfrac{7^2}{2.9}\)+\(\dfrac{7^2}{9.16}\)+\(\dfrac{7^2}{16.23}\)+....+\(\dfrac{7^2}{65.72}\)
\(\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+\dfrac{7^2}{16.23}+...+\dfrac{7^2}{65.72}\)
\(=7^2\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right)\)
\(=7^2\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)
\(=7^2\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)
\(=49\left(\dfrac{35}{72}\right)\)
\(=\dfrac{1715}{72}\)
\(l=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+\dfrac{7^2}{16.23}+...+\dfrac{7^2}{65.72}\)
\(=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+\dfrac{7}{16.23}+...+\dfrac{7}{65.72}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)=7\left(\dfrac{36}{72}-\dfrac{1}{72}\right)=7.\dfrac{35}{72}=\dfrac{245}{72}\)
Hãy tính \(\dfrac{C}{D}\). Biết C= \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\) và D= \(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
=> 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
Cho \(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+................+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\) và \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+..........+\dfrac{48}{2}+\dfrac{49}{1}\)
Tính \(\dfrac{S}{P}\)
Help me!!!!!!!!!!!
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)
Cho \(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50};B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
Tính giá trị của \(\dfrac{A}{B}\)
\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)
\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)
\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)
\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)
Tính: \(\dfrac{1}{3.10}+\dfrac{1}{10.17}+\dfrac{1}{17.24}+...+\dfrac{1}{73.80}-\dfrac{1}{2.9}-\dfrac{1}{9.16}-\dfrac{1}{16.23}-\dfrac{1}{23.30}+\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{3.6}+...+\dfrac{1}{97.99}-\dfrac{1}{98.100}\)
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
Cho P= \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\) và \(Q\)=\(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{47}{3}+\dfrac{48}{2}+\dfrac{49}{1}\)
Hãy tính \(\dfrac{P}{Q}\)
HELP ME!!!
\(Q=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{47}{3}+\dfrac{48}{2}+\dfrac{49}{1}\\ =\dfrac{1}{49}+1+\dfrac{2}{48}+1+\dfrac{3}{47}+1+...+\dfrac{47}{3}+1+\dfrac{48}{2}+1+1\\ =\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{3}+\dfrac{50}{2}+\dfrac{50}{50}\\ =50\cdot\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+...+\dfrac{1}{3}+\dfrac{1}{2}\right)\\ =50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}}{50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)