Violympic toán 6

NS

1.thực hiện phép tính

a.\(\frac{-1}{12}-\left(2^5_8-\frac{1}{3}\right)\)

b.-1,75-\(\left(\frac{-1}{9}-2^1_{18}\right)\)

c.\(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)

d.\(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)

e.\(\left(8^5_{11}+3^5_8\right)-3^5_{11}\)

f.\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6^5_9:\left(\frac{-1}{7}\right)\)

g.\(\frac{-1}{4}.13^9_{11}-0,25.6^2_{11}\)

h.\(5^{27}_5+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)

i.\(\frac{3}{8}.27^1_5-51^1_5.\frac{3}{8}+19\)

NT
2 tháng 7 2020 lúc 10:00

a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)

\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)

\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)

\(=-\frac{19}{8}\)

b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)

\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)

\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)

\(=\frac{5}{12}\)

c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)

\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)

\(=-\frac{43}{30}\)

d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)

\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)

\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)

\(=\frac{3}{20}\)

e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)

\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)

\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)

f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)

\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)

g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)

\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)

\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)

h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)

\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)

\(=\frac{78149}{6210}\)

i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)

\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)

\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)

\(=\frac{55}{4}\)

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