Tính \(\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-\dfrac{1}{95\cdot93}-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)
5 ps thôi nha ae
\(\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-\dfrac{1}{95\cdot93}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)
\(=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=\dfrac{1-48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)
Tính giá trị biểu thức:
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7} +.....................+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+..........+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.......+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...........+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+...........+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.............+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+..........+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{100}{2}\)
\(=50\)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+....+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+.....+\dfrac{1}{49.51}\right)}=\dfrac{\dfrac{100}{99}+\dfrac{100}{3.97}+....+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+....+\dfrac{1}{49.51}\right)}=\dfrac{100}{2}=50\)
\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-\frac{1}{95\cdot93}-....-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)
Đặt \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(A=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)
\(A=\frac{1}{99.97}-\left(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\right)\)
\(A=\frac{1}{99.97}-\left(\frac{1}{2}.\left(1-\frac{1}{97}\right)\right)=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}=\frac{1}{99.97}-\frac{48}{97}=-\frac{4751}{9603}\)
Tính giá trị của biểu thức \(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-\frac{1}{95\cdot93}-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)
Tính D = \(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-\frac{1}{95\cdot93}-....-\frac{1}{3\cdot1}\)
bn tách 1/ 97 .95 = 1/2 . ( 1/95 -1/97) nha! rồi sử dụng phương pháp khử liên tiếp !
Tính nhanh
\(\dfrac{8}{1\cdot3}-\dfrac{16}{3.5}+\dfrac{24}{5\cdot7}-\dfrac{32}{7\cdot9}+\dfrac{40}{9\cdot11}-...+\dfrac{384}{95\cdot97}-\dfrac{392}{97\cdot99}\)
Tính nhanh \(\dfrac{8}{1\cdot3}-\dfrac{16}{3.5}+\dfrac{24}{5\cdot7}-\dfrac{32}{7\cdot9}+\dfrac{40}{9\cdot11}-...+\dfrac{384}{95\cdot97}-\dfrac{392}{97\cdot99}\)
Tính\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-...-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)
bây giờ mìh ban rồi, mìh chỉ có thể chỉ cho bn cách làm thôi
dat bieu thuc la A
2A=2*(...)
2A=2/...-2/...
2A=(1/99-1/97)-(1/97-1/95)-...
2A=1/99-1=-98/99
A=...=-49/99
DUYỆT NHÉ
mìh cũng ko chắc chắn lắm đâu đấy nhé
Tính
A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\frac{1}{5\cdot95}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}\)