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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, \(\left(3\frac{1}{2}-2x\right).1\frac{1}{3}=7\frac{1}{3}\)
\(\frac{7}{2}-2x=\frac{22}{3}:\frac{4}{3}\)
\(\frac{7}{2}-2x=\frac{11}{2}\)
\(2x=\frac{7}{2}-\frac{11}{2}\)
\(2x=-2\)
\(\Rightarrow x=-2:2\)
\(x=-1\)
bn ơi ko thì bn làm từng phép một cũng đc nhé các bn giải hộ mình mai mình phải nộp rồi
1)
a) \(-\frac{9}{34}:\frac{17}{4}\)
\(=-\frac{18}{289}.\)
b) \(1\frac{1}{2}.\frac{1}{24}\)
\(=\frac{3}{2}.\frac{1}{24}\)
\(=\frac{1}{16}.\)
c) \(-\frac{5}{2}:\frac{3}{4}\)
\(=-\frac{10}{3}.\)
d) \(4\frac{1}{5}:\left(-2\frac{4}{5}\right)\)
\(=\frac{21}{5}:\left(-\frac{14}{5}\right)\)
\(=-\frac{3}{2}.\)
Mấy câu sau bạn đăng ríu rít quá khó nhìn lắm.
Chúc bạn học tốt!
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
\(\left|\frac{5}{-4}\right|-\left|\frac{1}{-3}\right|+-\frac{5}{6}-4\frac{1}{2}\)
\(=\left|-\frac{5}{4}\right|-\left|\frac{-1}{3}\right|+\frac{-5}{6}-\frac{9}{2}\)
\(=\frac{5}{4}-\frac{1}{3}+\frac{-5}{6}-\frac{9}{2}=-\frac{53}{12}\)
\(\frac{5}{8}-\left|-\frac{1}{12}\right|-3\frac{1}{4}+\left|-\frac{5}{6}\right|\)
\(=\frac{5}{8}-\frac{1}{12}-\frac{13}{4}+\frac{5}{6}=-\frac{15}{8}\)
\(\frac{3}{-7}+\left|-\frac{5}{12}\right|+3\frac{1}{4}+\left|-\frac{5}{6}\right|\)
\(=\frac{-3}{7}+\frac{5}{12}+\frac{13}{4}+\frac{5}{6}=\frac{57}{14}\)
\(1\frac{3}{5}-\left|\frac{1}{-4}\right|+\frac{2}{-3}-\left|-\frac{1}{2}\right|\)
\(=\frac{8}{5}-\left|\frac{-1}{4}\right|+\frac{-2}{3}-\frac{1}{2}\)
\(=\frac{8}{5}-\frac{1}{4}+\frac{-2}{3}-\frac{1}{2}\)
\(=\frac{27}{20}+\frac{-7}{6}=\frac{27}{20}-\frac{7}{6}=\frac{11}{60}\)