tính
\(P=\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1}.99+\frac{1}{3}.97+\frac{1}{5}.95+...+\frac{1}{97}.3+\frac{1}{99}.1}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\frac{1}{5\cdot99}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.99}+...+\frac{1}{99.1}}\)
\(=\frac{\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)
\(=\frac{100}{2}=50\)
\(H=\frac{\left(1+97\right)\left(1+\frac{97}{2}\right)\left(1+\frac{97}{3}\right)\left(1+\frac{97}{4}\right)+...+\left(1+\frac{97}{99}\right)}{\left(1+99\right)\left(1+\frac{99}{2}\right)\left(1+\frac{99}{3}\right)\left(1+\frac{99}{4}\right)+...+\left(1+\frac{99}{97}\right)}\)
Tính \(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{99.1}}\)
Tính \(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
6 ở đâu hả https://olm.vn/thanhvien/aihaibara0
Tính
A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\frac{1}{5\cdot95}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}\)
\(A=\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\frac{1}{5\cdot95}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}\)
Giúp mình giải bài này với
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 !
\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-\frac{1}{95\cdot93}-....-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)
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:
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+.....\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+\frac{1}{97\times3}+\frac{1}{99\times1}}\)
Có khó không giúp mình với chiều nay cần gấp
Ta xét riêng tử số:
\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+......+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{1\times99}+\frac{100}{3\times97}+\frac{100}{5\times95}+......+\frac{100}{49\times51}\)
\(=100\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)
Bây giờ xét đến mẫu số:
\(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=\frac{2}{1\times99}+\frac{2}{3\times97}+\frac{2}{5\times95}+......+\frac{2}{49\times51}\)
\(=2\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)
Vậy giá trị của biểu thức là: \(\frac{100}{2}=50\)
Tích giá trị các biểu thức:
a) A = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
b) B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
a) Đặt B = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}\)
\(=100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right)\)
Đặt C = \(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\)
\(=\left(\frac{1}{1.99}+\frac{1}{99.1}\right)+\left(\frac{1}{3.97}+\frac{1}{97.3}\right)+...+\left(\frac{1}{49.51}+\frac{1}{51.49}\right)\)
\(=2\cdot\frac{1}{1.99}+2\cdot\frac{1}{3.97}+...+2\cdot\frac{1}{49.51}\)
\(=2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)\)
Thay B và C vào A
\(\Rightarrow A=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}=\frac{100}{2}=50\)
b) Đặt E = \(\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}\)
\(=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\)
\(=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Thay E vào B
\(\Rightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)
a,
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}}\)
\(A=\frac{\left[1+\frac{1}{99}\right]+\left[\frac{1}{3}+\frac{1}{97}\right]+...+\left[\frac{1}{49}+\frac{1}{51}\right]}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}\)
\(A=\frac{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{99.1}}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}\)
\(A=\frac{100\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}=\frac{100}{2}=50\)
b, Ta có:
\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\left[1+\frac{98}{2}\right]+\left[1+\frac{97}{3}\right]+...+\left[1+\frac{1}{99}\right]+1\)
\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}=100\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right]\)
Thế vào:
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]}=\frac{1}{100}\)