Tính A=\(\dfrac{0,5}{3}+\dfrac{0,5}{6}+\dfrac{0,5}{10}+\dfrac{0,5}{15}+...+\dfrac{0,5}{1275}+\dfrac{0,5}{1326}\)
K
Khách
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Những câu hỏi liên quan
6 tháng 10 2021
\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)
GH
Gia Hân
VIP
9 tháng 7 2023
A = \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\) . \(\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . 0
A = 0*
*Vì số nào nhân với 0 cũng bằng 0 nên không cần tính kết quả của phép tính\(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\)
ND
Nguyễn Đăng Nhân
CTVHS
10 tháng 10 2023
a) \(\left[\left(-2,7\right)^4\right]^5-\left[\left(-2,7\right)^2\right]^{20}\)
\(=\left(-2,7\right)^{20}-\left(-2,7\right)^{20}\)
\(=0\)
b) \(\left(-0,5\right)^5:\left(-0,5\right)^3-\left(\dfrac{17}{2}\right)^7:\left(\dfrac{17}{2}\right)^6\)
\(=\left(-0,5\right)^2-\dfrac{17}{2}\)
\(=0,25-\dfrac{17}{2}\)
\(=-8,25\)
c) \(\left(8^{14}:4^{12}\right):\left(16^6:8^2\right)\)
\(=8^{14}:4^{12}:16^6\cdot8^2\)
\(=2^{48}:2^{24}:2^{24}\)
\(=0\)
TT
18 tháng 4 2021
a) \(\dfrac{5}{9}:\left(\dfrac{13}{7}+\dfrac{13}{9}\right)-\dfrac{5}{3}\)(chỗ này mk lười chép lại đề)
=\(\dfrac{5}{9}:\dfrac{208}{63}-\dfrac{5}{3}\)
=\(\dfrac{5}{9}.\dfrac{63}{208}-\dfrac{5}{3}\)
=\(\dfrac{5.63}{9.208}-\dfrac{5}{3}\)
=\(\dfrac{5.7}{1.208}-\dfrac{5}{3}\)
=\(\dfrac{36}{208}-\dfrac{5}{3}\)
=\(\dfrac{108}{624}-\dfrac{1040}{624}\)
=\(\dfrac{-932}{624}\)
=\(\dfrac{233}{156}\)
còn câu b mk chưa học nên mk chịu
18 tháng 4 2021
Giải:
5/9:13/7+5/9:13/9 -1 2/3
=5/9.7/13+5/9.9/13-5/3
=5/9.(7/13+9/13)-5/3
=5/9.16/13-5/3
=80/117-5/3
=-115/117
4 2/5 : 0,5% -1 3/7 .14% +(-0,5)
=22/5:1/200-10/7.7/50 +(-1/2)
=880-1/5-1/2
=8793/10
TL
1
19 tháng 5 2022
\(1\dfrac{1}{6}.2^2-0,5:\dfrac{3}{10}+\dfrac{1}{3}\)
\(=\dfrac{7}{6}.4-\dfrac{1}{2}.\dfrac{10}{3}+\dfrac{1}{3}\)
\(=\dfrac{7}{3}.2-\dfrac{1}{1}.\dfrac{5}{3}+\dfrac{1}{3}\)
\(=\dfrac{14}{3}-\dfrac{5}{3}+\dfrac{1}{3}\)
\(=\dfrac{14-5+1}{3}\)
\(=\dfrac{10}{3}\)
22 tháng 12 2021
\(\Leftrightarrow\left(x-0.5\right)\cdot\dfrac{-4}{x-0.5}=-1\cdot\left(-4\right)\)
=>-4=4(loại)
KH
8 tháng 6 2021
\(1\dfrac{13}{15}\cdot\left(0,5\right)^2+3\cdot\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4}+3\cdot\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{15}+3\cdot\dfrac{37}{20}\cdot\dfrac{24}{47}\)
\(=\dfrac{7}{15}+\dfrac{666}{235}=\dfrac{2327}{705}\)
9 tháng 6 2021
\(1\dfrac{13}{15}.\left(0,5\right)^2+3.\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.\dfrac{1}{4}+3.\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{7}{15}+3.\dfrac{37}{20}:\dfrac{47}{24}\)
\(=\dfrac{7}{15}+\dfrac{666}{235}\)
\(=\dfrac{2327}{705}\)
ND
1
14 tháng 4 2017
\(M=1\dfrac{1}{10}\cdot\dfrac{15}{19}-\dfrac{22}{38}\cdot0,5+110\%\cdot\dfrac{9}{19}\)
\(=\dfrac{11}{10}\cdot\dfrac{15}{19}-\dfrac{11}{19}\cdot\dfrac{1}{2}+\dfrac{11}{10}\cdot\dfrac{9}{19}\)
\(=\dfrac{15}{10}\cdot\dfrac{11}{19}-\dfrac{11}{19}\cdot\dfrac{1}{2}+\dfrac{9}{10}\cdot\dfrac{11}{19}\)
\(=\dfrac{11}{19}\left(\dfrac{15}{10}-\dfrac{1}{2}+\dfrac{9}{10}\right)\\ =\dfrac{11}{19}\cdot\dfrac{19}{10}\\ =\dfrac{11}{10}\)
Ta có:
\(A=\dfrac{0,5}{3}+\dfrac{0,5}{6}+\dfrac{0,5}{10}+...+\dfrac{0,5}{1275}+\dfrac{0,5}{1326}\)
\(\Rightarrow A=0,5\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{1}{1275}+\dfrac{1}{1326}\right)\)
\(\Rightarrow A=0,5.2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{10}+....+\dfrac{1}{2550}+\dfrac{1}{2652}\right)\)
\(\Rightarrow A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{10}+....+\dfrac{1}{2550}+\dfrac{1}{2652}\)
\(\Rightarrow A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{50.51}+\dfrac{1}{51.52}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{50}-\dfrac{1}{51}+\dfrac{1}{51}-\dfrac{1}{52}\)\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{26}{52}-\dfrac{1}{52}=\dfrac{25}{52}\)