\(\frac{13}{4}\)*1.13*1.99
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Q=\(\frac{3+1+\frac{3}{5}+...+\frac{3}{99}}{\left(\frac{1}{1.99}+\frac{1}{99.1}\right)+\left(\frac{1}{3.97}+\frac{1}{97.3}\right)+...+\left(\frac{1}{49.51}+\frac{1}{51.49}\right)}\)
Q=\(\frac{\frac{3}{1}+\frac{3}{3}+\frac{3}{5}+...+\frac{3}{99}}{\frac{2}{1.99}+\frac{2}{3.97}+...+\frac{2}{49.51}}\)
Q=\(50.\frac{3\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)}{50\left(\frac{2}{1.99}+\frac{2}{3.97}+...+\frac{2}{49.51}\right)}\)
Q=\(50.3.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}\)
Q=\(150.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{99+1}{1.99}+\frac{97+3}{3.97}+...+\frac{51+49}{49.51}}\)
Q=150\(.\frac{\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}\)
Q=\(150.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}\)
Q=150.1
Q=150
\(Q=\frac{4+\frac{3}{5}+...+\frac{3}{95}+\frac{3}{97}+\frac{3}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{95.5}+\frac{1}{97.3}+\frac{1}{99.1}}\)
=> \(Q=\frac{100\left(\frac{3}{1}+\frac{3}{3}+\frac{3}{5}+...+\frac{3}{95}+\frac{3}{97}+\frac{3}{99}\right)}{100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{95.5}+\frac{1}{97.3}+\frac{1}{99.1}\right)}\)
=> \(Q=\frac{100.3\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{\frac{1+99}{1.99}+\frac{3+97}{3.97}+\frac{5+95}{5.95}+...+\frac{95+5}{95.5}+\frac{97+3}{97.3}+\frac{99+1}{99.1}}\)
=> \(Q=\frac{300\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{95}+\frac{1}{5}\right)+\left(\frac{1}{97}+\frac{1}{3}\right)+\left(\frac{1}{99}+\frac{1}{1}\right)}\)
=> \(Q=\frac{300\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{2\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}\)
=> \(Q=\frac{300}{2}=150\)
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}=\frac{\left[1+\frac{1}{99}\right]+\left[\frac{1}{3}+\frac{1}{97}\right]+...+\left[\frac{1}{49}+\frac{1}{51}\right]}{2\left[\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{49\cdot51}\right]}\)
\(=\frac{\frac{100}{1\cdot99}+\frac{100}{3\cdot97}+...+\frac{100}{49.51}}{2\left[\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{49.51}\right]}=\frac{100\left[\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{49.51}\right]}{2\left[\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{49.51}\right]}=\frac{100}{2}=50\)
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 \(A=50\)
\(\frac{13}{4}\)* 1,13 * 1,99
= 3,25 * 1,13 * 1,99
= 3,6725 * 1,99
= 7,308275
con chó Nguyễn Hoài Nam óc chó