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6 tháng 1 2023
A. 2.9 +6 -4 = 20
B. 8.25 - 6.9 + (-12) = 134
C. -962 + 1548 = 586
D. -85.(36+54) = -85 x 100 = -8500
F. 26.37 - 26.56 - 56.37 + 56.26 =26.37 - 56.37 = 37.(26 - 56)
=37. (-30) = -1100
E. 130 - ( 27 - (36 + 128).2))3
= 130 - ( 27 - 328).3
= 130 - (-301).3
= 130 - (-903)
= 1033
9 tháng 3 2018
a. \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{17.20}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{17}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
b. \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{4}-\dfrac{1}{10}\)
\(=\dfrac{3}{20}\)
c. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)
\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)
\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=4\left(1-\dfrac{1}{49}\right)\)
\(=4.\dfrac{48}{49}\)
\(=\dfrac{192}{49}\)
18 tháng 1
a: \(\dfrac{16}{15}\cdot\dfrac{-5}{14}\cdot\dfrac{54}{24}\cdot\dfrac{56}{21}\)
\(=\dfrac{16}{24}\cdot\dfrac{-5}{15}\cdot\dfrac{54}{21}\cdot\dfrac{56}{14}\)
\(=\dfrac{2}{3}\cdot\dfrac{-1}{3}\cdot\dfrac{18}{7}\cdot4\)
\(=\dfrac{-2\cdot2\cdot4}{7}=-\dfrac{16}{7}\)
b: \(\dfrac{7}{3}\cdot\dfrac{-5}{2}\cdot\dfrac{15}{21}\cdot\dfrac{4}{-5}\)
\(=\dfrac{7}{3}\cdot\dfrac{5}{7}\cdot\dfrac{5}{2}\cdot\dfrac{4}{5}\)
\(=\dfrac{5}{3}\cdot\dfrac{4}{2}=\dfrac{5}{3}\cdot2=\dfrac{10}{3}\)
c: \(\left(-\dfrac{2}{5}+\dfrac{1}{3}\right)\left(\dfrac{3}{2}-\dfrac{3}{7}\right)\)
\(=\dfrac{-2\cdot5+3}{15}\cdot\dfrac{3\cdot7-3\cdot2}{14}\)
\(=\dfrac{-7}{15}\cdot\dfrac{15}{14}=\dfrac{-7}{14}=-\dfrac{1}{2}\)
d: \(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\left(5-\dfrac{1}{4}\right)\)
\(=\dfrac{3-2}{6}\cdot\dfrac{20-1}{4}\)
\(=\dfrac{1}{6}\cdot\dfrac{19}{4}=\dfrac{19}{24}\)
e: \(\left(\dfrac{4}{5}\right)^2=\dfrac{4^2}{5^2}=\dfrac{16}{25}\)
3 tháng 2 2022
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
Ta có : A = 1 + 2 + 22 + 23 + ....... + 256
=> 2A = 2 + 22 + 23 + ....... + 257
=> 2A - A = 257 - 1
=> A = 257 - 1
Vậy A = 257 - 1