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Đặt \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(A=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)
\(A=\frac{1}{99.97}-\left(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\right)\)
\(A=\frac{1}{99.97}-\left(\frac{1}{2}.\left(1-\frac{1}{97}\right)\right)=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}=\frac{1}{99.97}-\frac{48}{97}=-\frac{4751}{9603}\)
\(=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=\dfrac{1-48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)
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
Sửa đề: \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{3.1}\)
\(=\dfrac{1}{97.99}-\left(\dfrac{1}{1.3}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(\dfrac{2}{1.3}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{97.99}\right)-\dfrac{1}{2}.\dfrac{96}{97}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{48}{97}\)
.........................
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.99}+...+\frac{1}{99.1}}\)
\(=\frac{\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{100}{2}=50\)
bn tách 1/ 97 .95 = 1/2 . ( 1/95 -1/97) nha! rồi sử dụng phương pháp khử liên tiếp !
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