Tính: A= \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{24}\)+\(\dfrac{1}{48}\)+\(\dfrac{1}{96}\)
Tính tổng sau đây bằng cách nhanh nhất
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\)
GIÚP MÌNH VỚI NHÉ MÌNH ĐAG CẦN GẤP.CHO MÌNH GỬI LỜI CẢM ƠN TRC NHÉ!
=\(\dfrac{64}{192}+\dfrac{32}{192}+\dfrac{16}{192}+\dfrac{8}{192}+\dfrac{4}{192}+\dfrac{2}{192}+\dfrac{1}{192}\)
= \(\dfrac{127}{192}\)
Tính nhanh:
a, \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
b, \(\left(-\dfrac{1}{4}+\dfrac{7}{35}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{48}{49}\right)\)
a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
A = \(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}\)
\(\dfrac{-1}{7}-\dfrac{1}{8}\)
\(\dfrac{-15}{48}-\dfrac{1}{12}\)
\(\dfrac{-3}{4}-\dfrac{4}{5}\)
\(\dfrac{3}{4}+\dfrac{-5}{6}-\dfrac{11}{12}\)
\(\dfrac{-1}{7}-\dfrac{1}{8}=\dfrac{-8}{56}-\dfrac{7}{56}=\dfrac{-15}{56}\\ \dfrac{-15}{48}-\dfrac{1}{12}=\dfrac{-5}{16}-\dfrac{1}{12}=\dfrac{-15}{48}-\dfrac{4}{48}=\dfrac{-19}{48}\\ \dfrac{-3}{4}-\dfrac{4}{5}=\dfrac{-15}{20}-\dfrac{16}{20}=\dfrac{-31}{20}\\ \dfrac{3}{4}+\dfrac{-5}{6}-\dfrac{11}{12}=\dfrac{9}{12}-\dfrac{10}{12}-\dfrac{11}{12}=\dfrac{-12}{12}=-1\)
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: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
Bài 1. Tính nhanh giá trị của biểu thức sau:
a, E = \(\dfrac{1}{2}\)+ \(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\)+ \(\dfrac{1}{24}\)+ \(\dfrac{1}{8}\)+ \(\dfrac{1}{2}\)+\(\dfrac{1}{12}\)
\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)
\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)
\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)
\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)
\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)
\(E=\dfrac{3}{2}+\dfrac{1}{4}\)
\(E=\dfrac{6}{4}+\dfrac{1}{4}\)
\(E=\dfrac{7}{4}\)
TÍNH NHANH:
A=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)
B=\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{95}\)
C=\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+...+\dfrac{1}{9800}\)
*D=\(\dfrac{2}{3.5}+\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{13}{19.32}+\dfrac{25}{32.57}+\dfrac{30}{57.87}\)
\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(B=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+..+\dfrac{1}{195}\) ( là 195 ms đúng ! )
\(B=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{13\cdot15}\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{15}\right)=\dfrac{1}{2}\cdot\dfrac{14}{15}=\dfrac{7}{15}\)
\(C=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{98\cdot100}\)
Rồi làm tương tự cân b nha!
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}\)
\(+\dfrac{1}{57}-\dfrac{1}{87}\)
\(D=\dfrac{1}{3}-\dfrac{1}{87}=\dfrac{28}{87}\)
Tính
a)\(\dfrac{-3}{14}\)+\(\dfrac{5}{8}\)-\(\dfrac{1}{2}\) b)\(\dfrac{5}{12}\)-\(\dfrac{24}{5}\).\(\dfrac{1}{6}\)
c)\(\dfrac{5}{12}\)-\(\dfrac{24}{5}\).\(\dfrac{1}{6}\) d)Tìm một số biết:
-24 là \(\dfrac{3}{8}\)của số đó
Giúp mik nhanh nha mik cần gấp
a: \(=\dfrac{-12}{56}+\dfrac{35}{56}-\dfrac{28}{56}=-\dfrac{5}{56}\)
b: \(=\dfrac{5}{12}-\dfrac{4}{5}=\dfrac{25-48}{60}=\dfrac{-23}{60}\)
d: SỐ cần tìm là:
-24:3/8=-24x8:3=-64
a \(\dfrac{-5}{56}\)
b \(\dfrac{-23}{60}\)
c \(\dfrac{-23}{60}\)
d \(\dfrac{-1}{64}\)
\(\dfrac{1}{4}+\dfrac{1}{2}\times\dfrac{1}{3}\)
\(=\dfrac{1}{4}+\dfrac{1}{6}\)
\(=\dfrac{10}{24}=\dfrac{5}{12}\)