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KK
14 tháng 9 2021
a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 7
b; 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)
= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))
= 1 + 1
= 2
13 tháng 8 2019
a) A = \(9\frac{3}{8}-\left(2\frac{3}{5}+2\frac{3}{8}\right)=9\frac{3}{8}-2\frac{3}{5}-2\frac{3}{8}=\left(9\frac{3}{8}-2\frac{3}{8}\right)-2\frac{3}{5}=7-\frac{13}{5}=\frac{22}{5}\)
b) B = \(\left(15\frac{3}{5}+5\frac{3}{4}\right)-8\frac{3}{5}=15\frac{3}{5}+5\frac{3}{4}-8\frac{3}{5}=\left(15\frac{3}{5}-8\frac{3}{5}\right)+5\frac{3}{4}=7+\frac{23}{4}=\frac{51}{4}\)
c) C = \(17\frac{1}{4}-\left(2\frac{3}{7}+7\frac{1}{4}\right)=17\frac{1}{4}-2\frac{3}{7}-7\frac{1}{4}=\left(17\frac{1}{4}-7\frac{1}{4}\right)-2\frac{3}{7}=10-\frac{17}{7}=\frac{53}{7}\)
d) D = \(\left(11\frac{5}{17}+3\frac{5}{7}\right)-4\frac{5}{17}=11\frac{5}{17}+3\frac{5}{7}-4\frac{5}{17}=\left(11\frac{5}{17}-4\frac{5}{17}\right)+3\frac{5}{7}=7+\frac{26}{7}=\frac{75}{7}\)
27 tháng 7 2021
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
\(A=\left(\frac{5}{3}-\frac{3}{7}+9\right)-\left(2+\frac{5}{7}-\frac{2}{3}\right)+\left(\frac{8}{7}-\frac{4}{3}-10\right)\)
\(=\frac{5}{3}-\frac{3}{7}+9-2-\frac{5}{7}+\frac{2}{3}+\frac{8}{7}-\frac{4}{3}-10\)
\(=\left(\frac{5}{3}+\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{7}+\frac{5}{7}-\frac{8}{7}\right)+\left(9-2-10\right)\)
\(=1-0-3\)
\(=-2\)
5/3-3/7+9-2-5/7+2/3+8/7-4/3-10
(5/3-4/3+2/3)+(8/7-3/7-5/7)+(9-2-10)
1+0-3=-2