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a: \(=\dfrac{-5}{8}+\dfrac{7}{12}=\dfrac{-15}{24}+\dfrac{14}{24}=\dfrac{-1}{24}\)
b: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{11}{12}=\dfrac{9}{12}-\dfrac{10}{12}+\dfrac{11}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)
c: \(=\dfrac{6}{36}-\dfrac{1}{36}=\dfrac{5}{36}\)
d: \(=\dfrac{5}{12}+\dfrac{5}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)
e: \(=\dfrac{-8}{56}-\dfrac{7}{56}=\dfrac{-15}{56}\)
f: \(=\dfrac{-5}{15}+\dfrac{3}{25}=\dfrac{-25}{75}+\dfrac{9}{75}=\dfrac{-16}{75}\)
A = - 2 / 5 + ( - 2 / 5 + 2 )
A = - 2 / 5 + 8 / 5
A = 6 / 5
B = ( - 5 / 24 + 0, 75 + 7 / 12 ) : - 17/8
B = 9 / 8 : - 17/8
B = -9 / 17
C = 3 / 7 + ( - 1 / 5 + - 3 / 7 )
C = 3 / 7 - 1 / 5 - 3 / 7
C = - 1 / 5
D = ( 6 - 14 / 5 ) . 25 / 8 - 8 / 5 : 1 / 4
D = 16 / 5 . 25 / 8 - 8 / 5 . 4
D = 10 - 6 , 4
D = 3,6
\(A=-\frac{2}{5}+\left[\left(-\frac{2}{5}\right)+2\right]=\left[\left(-\frac{2}{5}\right)+\left(-\frac{2}{5}\right)\right]+2=\left(-\frac{4}{5}\right)+2=\left(-\frac{4}{5}\right)+\frac{10}{5}=\frac{6}{5}\)\(\frac{10}{5}=\frac{6}{5}\)
\(B=\left[\left(-\frac{5}{24}\right)+0,75+\frac{7}{12}\right]:\left(\frac{-17}{8}\right)\)\(=\left[\left(-\frac{5}{24}\right)+\frac{3}{4}+\frac{7}{12}\right].\frac{-8}{17}\)
\(=\left[\left(-\frac{5}{24}\right)+\frac{18}{24}+\frac{14}{24}\right].\frac{-8}{17}\)\(=\frac{27}{24}.\frac{-8}{17}=\frac{9}{8}.\frac{-8}{17}=\frac{9.\left(-8\right)}{8.17}=\frac{-9}{17}\)
\(C=\frac{3}{7}+\left[\left(-\frac{1}{5}\right)+\left(-\frac{3}{7}\right)\right]=\left[\frac{3}{7}+\left(-\frac{3}{7}\right)\right]+\left(-\frac{1}{5}\right)=-\frac{1}{5}\)
\(D=\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}:\frac{1}{4}=\left(\frac{30}{5}-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}.4\)
\(=\frac{16}{5}.\frac{25}{8}-\frac{32}{5}=\frac{16.25}{5.8}-\frac{32}{5}=10-\frac{32}{5}\)\(=\frac{50}{5}-\frac{32}{5}=\frac{18}{5}\)
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
Bài giải:
Câu 1: a, \(\left(-2\right).4.5.38.\left(-25\right)\)
\(=\left[\left(-2\right).5\right].\left[4.\left(-25\right)\right].38\)
\(=\left(-10\right).\left(-100\right).38\)
\(=1000.38=38000\)
b,\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}\)
\(=\left(\frac{1}{3}+\frac{3}{8}\right)-\frac{7}{12}\)
\(=\frac{17}{24}-\frac{7}{12}=\frac{1}{8}\)
c, \(\frac{-5}{8}.\frac{5}{12}+\frac{-5}{8}.\frac{7}{12}+2\frac{1}{8}\)
\(=\frac{-5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{17}{8}\)
\(=\frac{-5}{8}.1+\frac{17}{8}\)
\(=\frac{3}{2}\)
Câu 2: a, \(x-\frac{2}{5}=0,24\)
\(x-0,4=0,24\)
\(x=0,24+0,4\)
\(\Rightarrow x=0,64\left(\frac{16}{25}\right)\)
b,\(\frac{2}{3}.x+\frac{1}{12}=\frac{1}{10}\)
\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{12}\)
\(\frac{2}{3}.x=\frac{1}{60}\)
\(x=\frac{1}{60}:\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{40}\)
c, \(\left(3\frac{1}{2}-2x\right).1\frac{1}{3}=7\frac{1}{3}\)
\(\frac{7}{2}-2x=\frac{22}{3}:\frac{4}{3}\)
\(\frac{7}{2}-2x=\frac{11}{2}\)
\(2x=\frac{7}{2}-\frac{11}{2}\)
\(2x=-2\)
\(\Rightarrow x=-2:2\)
\(x=-1\)
\(A=\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}=\dfrac{21}{36}+\dfrac{5}{12}.\dfrac{1}{6}-\dfrac{11}{36}=\dfrac{5}{18}+\dfrac{5}{12}.\dfrac{1}{6}=\dfrac{20}{72}+\dfrac{5}{72}=\dfrac{25}{72}\)
\(B=1\dfrac{3}{8}+\dfrac{1}{8}:\left(0,75-\dfrac{1}{2}\right)-25\%.\dfrac{1}{2}=\dfrac{11}{8}+\dfrac{1}{8}:\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{5}{4}+\dfrac{1}{2}=\dfrac{7}{4}\)
A. 25/72
B. 7/4