Đặt \(A=\dfrac{24\cdot135+3\cdot561\cdot8+4\cdot126\cdot6}{1+3+5+7+...+97+99-500}\)
\(=\dfrac{24\cdot822}{2000}=\dfrac{1233}{125}\)
Cho M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+..........+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..........+\dfrac{1}{100}}\) ; N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-.........-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+......+\dfrac{1}{500}}\)
Tìm tỉ số phần trăm của M và N
-1+3-5+7+...+97-99
-1+3-5+7+...+97-99
tính Q = \(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
1+2+(-3)+4+5+(-6)...+97+98+(-99)
tìm x biết \(\dfrac{1}{x}\)-\(\dfrac{1}{9999}\)=\(\dfrac{1}{\begin{matrix}1&\times&3\end{matrix}}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{97\times99}\)
cho M=\(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\)
N=\(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\)
chứng minh rằng: M<\(\dfrac{1}{10}\)
Em cần gấp câu trả lời cho bài toán này, mong đc mn giúp đỡ (nếu được xin trả lời trước 12h ngày 10/5 giúp em ạ). Cảm ơn mn.
a, \(\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}-\dfrac{4}{3}\)
b, \(\dfrac{4}{19}-\dfrac{-3}{7}+\dfrac{-3}{7}.\dfrac{15}{19}+\dfrac{5}{7}\)
c, \(\dfrac{5}{9}.\dfrac{7}{13}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}.\dfrac{3}{13}\)
Tính giá trị biểu thức
a, \(\dfrac{5.8-5.6}{10}\) b, \(\dfrac{\left(-4\right)^2}{5}\) B = \(\dfrac{3}{7}+\left(\dfrac{-1}{5}+\dfrac{-3}{7}\right)\) C = \(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\)
D = \(\left(\dfrac{-5}{24}+0,75+\dfrac{7}{12}\right):\left(-2\dfrac{1}{8}\right)\) E = \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\) F = \(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
