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Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
1) 4824 - 4824 : 24 - 12 = 4824 - 201 - 12 = 4623 - 12 = 4611
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
1) \(\left|\dfrac{-1}{7}+\dfrac{-4}{3}\right|+\dfrac{-5}{2}\)
\(=\left|\dfrac{-31}{21}\right|+\dfrac{-5}{2}\)
\(=\dfrac{31}{21}+\dfrac{-5}{2}=\dfrac{-43}{42}.\)
2) \(\dfrac{-3}{2}+\left|\dfrac{-4}{7}\right|+\left|\dfrac{-1}{5}\right|\)
\(=\dfrac{-3}{2}+\dfrac{4}{7}+\dfrac{1}{5}\)
\(=\dfrac{-51}{70}.\)
1) \(\left|\dfrac{-1}{7}+\dfrac{-4}{3}\right|+\dfrac{-5}{2}\)
= \(\left|\dfrac{-3}{21}+\dfrac{-28}{21}\right|+\dfrac{-5}{2}\)
= \(\left|\dfrac{-31}{21}\right|+\dfrac{-5}{2}\)
= \(\dfrac{31}{21}+\dfrac{-5}{2}\)
= \(\dfrac{62}{42}\)+\(\dfrac{-105}{42}\)
= \(\dfrac{-43}{42}\)