Violympic toán 6

H24

Bài 1 : Tính

1) \(\frac{4}{7}-\frac{1}{14}+\left|\frac{-5}{21}\right|\)

2) \(\left|\frac{-2}{3}\right|-\frac{1}{2}+3\)

3) \(\left|\frac{7}{-4}\right|-\frac{5}{8}+\frac{-2}{3}\)

4) \(\frac{4}{5}+\left|\frac{-3}{2}\right|+\frac{1}{-4}\)

5) \(\left|\frac{-1}{4}\right|-3+\frac{3}{4}\)

6) \(\left|\frac{-1}{3}\right|-\frac{5}{4}+\frac{1}{5}\)

ND
8 tháng 8 2020 lúc 15:25

1) \(\frac{4}{7}-\frac{1}{14}+\left|\frac{-5}{21}\right|\)

\(=\frac{4}{7}-\frac{1}{14}+\frac{5}{21}\)

\(=\frac{24}{42}-\frac{3}{42}+\frac{10}{42}\)

\(=\frac{31}{42}\)

2) \(\left|\frac{-2}{3}\right|-\frac{1}{2}+3\)

\(=\frac{2}{3}-\frac{1}{2}+\frac{3}{1}\)

\(=\frac{4}{6}-\frac{3}{6}+\frac{18}{6}\)

\(=\frac{19}{6}\)

3) \(\left|\frac{7}{-4}\right|-\frac{5}{8}+\frac{-2}{3}\)

\(=\frac{7}{4}-\frac{5}{8}+\frac{-2}{3}\)

\(=\frac{42}{24}-\frac{15}{24}+\frac{-16}{24}\)

\(=\frac{11}{24}\)

4) \(\frac{4}{5}+\left|\frac{-3}{2}\right|+\frac{1}{-4}\)

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

\(=\frac{16}{20}+\frac{30}{20}+\frac{-5}{20}\)

\(=\frac{41}{20}\)

5) \(\left|\frac{-1}{4}\right|-3+\frac{3}{4}\)

\(=\frac{1}{4}-\frac{3}{1}+\frac{3}{4}\)

\(=\frac{1}{4}-\frac{12}{4}+\frac{3}{4}\)

\(=-2\)

6) \(\left|\frac{-1}{3}\right|-\frac{5}{4}+\frac{1}{5}\)

\(=\frac{1}{3}-\frac{5}{4}+\frac{1}{5}\)

\(=\frac{20}{60}-\frac{75}{60}+\frac{12}{60}\)

\(=\frac{-43}{60}\)

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