\(choA=\dfrac{1}{1x99}+\dfrac{1}{3x97}+\dfrac{1}{5x95}+...+\dfrac{1}{97x3}+\dfrac{1}{99x1}\)
\(B=1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}\)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1x99}+\dfrac{1}{3x97}+\dfrac{1}{5x95}+...+\dfrac{1}{97x3}+\dfrac{1}{99x1}}\)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1x99}+\dfrac{1}{3x97}+\dfrac{1}{5x95}...+\dfrac{1}{49x51}}\)Rút gọn biểu thức trên về tối giản
(1+1/3+1/5+1/7+...+1/97+1/99):(1/1x99+1/3x97+1/5x95+...+1/97x3+1/99x1)
Đặt \(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right):\left(\frac{1}{1.99}+\frac{1}{3.97}+....+\frac{1}{97.3}+\frac{1}{99.1}\right)\)
Đặt \(B=\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{97.3}+\frac{1}{99.1}\)
=> 100 x B = \(\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{97.3}+\frac{100}{99.1}=1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+...+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1\)
=> 100 x B = \(2.\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)\)
=> \(B=\frac{1}{50}.\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)\)
Khi đó A = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{50}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{\frac{1}{50}}=50\)
1+1/3+1/5+1/7+...+.1/97+1/99/1/1x99+1/3x97+1/5x95+...+1/97x3+1/99x1
A=\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\times99}+\dfrac{1}{3\times97}+\dfrac{1}{5\times95}+...+\dfrac{1}{97\times3}+\dfrac{1}{99\times1}}\)
Ta có: \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{100}{1\cdot99}+\dfrac{100}{3\cdot97}+\dfrac{100}{5\cdot95}+...+\dfrac{100}{97\cdot3}+\dfrac{100}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{1+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+\dfrac{1}{5}+\dfrac{1}{95}+...+\dfrac{1}{97}+\dfrac{1}{3}+\dfrac{1}{99}+1}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}\right)}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1}{2}\)
hay A=50
Tính giá trị của biểu thức:
\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.99}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{99}+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+....+\dfrac{1}{99}}\)
Ta có:
\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\dfrac{99+1}{1\cdot99}+\dfrac{97+3}{3\cdot97}+...+\dfrac{1+99}{99\cdot1}}{100}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{99}+1\right)}{100}}\)
\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}}=\dfrac{1}{\dfrac{2}{100}}=\dfrac{100}{2}=50\)
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+...+\dfrac{1}{99}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{1+\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{100}{100}+\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)
Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)
tính
1+ 1/3+ 1/5+1/7+...+1/97+1/99
1/1x99 + 1/ 3x97 + 1/5x95 +...+ 1/97x3 + 1/99x1
Người phán sử
Gọi biểu thức đó là A ta có :
\(A=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\)
\(A=2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(A=2-\frac{1}{99}\)
\(A=\frac{197}{99}\)
thế còn phần ở dưới đâu?đó là một bài đấy. hiểu chưa?????????????/
\(A=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\)
\(A=2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
A = 197/99
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)