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2 tháng 4 2017

Bài 1: Tính

\(\text{1)}\) \(\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{2}.\dfrac{1}{8}\)

\(=\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{8}.\dfrac{1}{2}\)

\(=\dfrac{5}{8}.\left(\dfrac{7}{30}-\dfrac{1}{2}\right)\)

\(=\dfrac{5}{8}.\dfrac{-4}{15}\)

\(=\dfrac{-1}{6}\)

\(\text{2)}\) \(\dfrac{21}{10}.\dfrac{3}{4}-\dfrac{21}{10}-\dfrac{3}{4}\)

\(=\dfrac{63}{40}-\dfrac{21}{10}-\dfrac{3}{4}\)

\(=\dfrac{-21}{40}-\dfrac{3}{4}\)

\(=\dfrac{-51}{40}\)

\(\text{3)}\) \(\dfrac{-4}{11}:\dfrac{-6}{11}\)

\(=\dfrac{-4}{11}.\dfrac{11}{-6}\)

\(=\dfrac{4}{6}\)

\(\text{4)}\) \(\dfrac{2}{7}.\dfrac{14}{3}-1\)

\(=\dfrac{4}{3}-1\)

\(=\dfrac{1}{3}\)

\(\text{5)}\) \(\dfrac{4}{7}:\left(\dfrac{1}{5}.\dfrac{4}{7}\right)\)

\(=\dfrac{4}{7}:\dfrac{1}{5}:\dfrac{4}{7}\)

\(=1:\dfrac{1}{5}\)

\(=5\)

\(\text{6)}\) \(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)

\(=3+\dfrac{35}{11}.\dfrac{121}{245}\)

\(=3+\dfrac{11}{7}\)

\(=3\dfrac{11}{7}=\dfrac{32}{7}\)

\(\text{7)}\) \(\left(\dfrac{4}{3}+\dfrac{8}{3}\right).\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)

\(=4.\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)

\(=4.\dfrac{1}{4}:\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)

\(=4.\dfrac{1}{4}:\dfrac{19}{5}\)

\(=1:\dfrac{19}{5}\)

\(=\dfrac{5}{19}\)

\(\text{8)}\) \(\left(\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{\dfrac{1}{9}}{\dfrac{1}{9}}\right):\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{2}{5}}-\dfrac{1}{6}\right)\)

\(=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{15}:\dfrac{2}{5}-\dfrac{1}{6}\right)\)

\(=1:\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)\)

\(=1:\left(\dfrac{2}{3}+1\right)\)

\(=1:\dfrac{5}{3}\)

\(=\dfrac{3}{5}\)
\(\text{9)}\)

\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\left[\dfrac{199}{75077}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\left[\dfrac{199}{6613}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\dfrac{13235}{13226}:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\dfrac{13235}{13226}:\left[\dfrac{3}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\dfrac{13235}{13226}:\left[\dfrac{3}{50}+\dfrac{9}{2}\right]\)

\(=\dfrac{13235}{13226}:\dfrac{114}{25}\)

\(=\dfrac{330875}{1507764}\)

2 tháng 4 2017

1) \(\frac{5}{8}.\frac{7}{3}-\frac{5}{2}.\frac{1}{8}=\frac{5}{8}.\frac{7}{3}-\frac{5}{8}.\frac{1}{2}=\frac{5}{8}\left(\frac{7}{3}-\frac{1}{2}\right)=\frac{5}{8}.\frac{11}{6}=\frac{55}{48}\)

2) \(\frac{21}{10}.\frac{3}{4}-\frac{21}{10}.\frac{3}{4}=\frac{21}{10}\left(\frac{3}{4}-\frac{3}{4}\right)=\frac{21}{10}.0=0\)

3) \(\frac{-4}{11}:\frac{-6}{11}=\frac{-4}{11}.\frac{-11}{6}=\frac{-4.\left(-11\right)}{11.6}=\frac{-4.\left(-1\right)}{1.6}=\frac{4}{6}=\frac{2}{3}\)

4)\(\frac{2}{7}.\frac{14}{3}-1=\frac{2.14}{7.3}-1=\frac{2.2}{1.3}-1=\frac{4}{3}-1=\frac{1}{3}\)

5)\(\frac{4}{7}:\left(\frac{1}{5}.\frac{4}{7}\right)=\frac{4}{7}:\frac{4}{35}=\frac{4}{7}.\frac{35}{4}=\frac{4.35}{7.4}=\frac{1.5}{1.1}=5\)

6) \(\frac{12}{7}.\frac{7}{4}+\frac{35}{11}:\frac{245}{121}=\frac{12.7}{7.4}+\frac{35}{11}.\frac{121}{245}=\frac{3.1}{1.1}+\frac{35}{11}.\frac{121}{245}=3+\frac{35}{11}.\frac{121}{245}=3+\frac{35.121}{11.245}=\frac{1.11}{1.7}=\frac{11}{7}\)

2 tháng 4 2017

bạn làm giùm mình kết bài con lại nhá

11 tháng 3 2021

a) \(\dfrac{\dfrac{3}{14}-\dfrac{3}{17}+\dfrac{3}{19}}{\dfrac{5}{19}+\dfrac{5}{14}-\dfrac{5}{17}}=\dfrac{3\left(\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{19}\right)}{5\left(\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{19}\right)}=\dfrac{3}{5}\)

 

11 tháng 3 2021

c) \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{193}}{\dfrac{3}{7}+\dfrac{3}{5}+\dfrac{3}{17}-\dfrac{3}{293}}=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{193}\right)}{3\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{193}\right)}=\dfrac{2}{3}\)

d) 

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{11}{33}-\dfrac{35}{40}\)

`=`\(\dfrac{1}{2}-\dfrac{5}{6}+\dfrac{1}{3}-\dfrac{7}{8}\)

`=`\(\dfrac{12}{24}-\dfrac{20}{24}+\dfrac{8}{24}-\dfrac{21}{24}\)

`= -21/24 = -7/8`

`b)`

\(\dfrac{2}{3}\cdot1\dfrac{3}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{2}{3}\cdot\dfrac{7}{4}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{7}{6}-\dfrac{8}{9}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(\dfrac{5}{18}-\dfrac{17}{51}-\dfrac{1}{5}\)

`=`\(-\dfrac{1}{18}-\dfrac{1}{5}=-\dfrac{23}{90}\)

`c)`

\(\dfrac{1}{2}\cdot2-2\dfrac{5}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(1-\dfrac{19}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(-\dfrac{12}{7}+\dfrac{6}{4}-\dfrac{10}{15}\)

`=`\(-\dfrac{3}{14}-\dfrac{10}{15}=-\dfrac{37}{42}\)

`d) `

\(\dfrac{1}{6}\cdot\dfrac{1}{11}+\dfrac{4}{11}\cdot\left(-\dfrac{1}{6}\right)+\dfrac{8}{11}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{6}{11}\)

`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1}{11}-\dfrac{4}{11}+\dfrac{8}{11}+\dfrac{6}{11}\right)\)

`=`\(\dfrac{1}{6}\cdot\left(\dfrac{1-4+8+6}{11}\right)\)

`=`\(\dfrac{1}{6}\cdot1=\dfrac{1}{6}\)

`e)`

\(-17\cdot\left(-23\right)+\left(-53\right)\cdot17+17\cdot14+17\cdot\left(-24\right)\)

`= 17*(23-53+14-24)`

`= 17*(-40)`

`= -680`

`f)`

\(-19\cdot218+\left(-82\right)\cdot19-533\cdot19+\left(-19\right)\cdot167\)

`= 19*(-218-82-533-167)`

`= 19*(-1000)`

`= -19000`

`g)`

\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{11}{44}+\dfrac{9}{16}\)

`=`\(\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{9}{16}\)

`=`\(\dfrac{31}{40}-\dfrac{1}{4}+\dfrac{9}{16}\)

`=`\(\dfrac{21}{40}+\dfrac{9}{16}=\dfrac{87}{80}\)

`h)`

\(\dfrac{4}{10}-1\dfrac{5}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{4}{10}-\dfrac{11}{6}\cdot2+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{4}{10}-\dfrac{11}{3}+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(-\dfrac{49}{15}+\dfrac{7}{8}-\dfrac{1}{9}\)

`=`\(-\dfrac{287}{120}-\dfrac{1}{9}=-\dfrac{901}{360}\)

`i )`

\(3\cdot\dfrac{1}{5}-\dfrac{2}{8}-\dfrac{12}{36}+\dfrac{15}{9}\)

`=`\(\dfrac{3}{5}-\dfrac{1}{4}-\dfrac{1}{3}+\dfrac{15}{9}\)

`=`\(\dfrac{7}{20}-\dfrac{1}{3}+\dfrac{15}{9}\)

`=`\(\dfrac{1}{60}+\dfrac{15}{9}=-\dfrac{33}{20}\)

`k)`

\(\dfrac{6}{8}\cdot3\dfrac{1}{2}+4\dfrac{2}{3}-\dfrac{11}{55}+\dfrac{17}{51}\)

`=`\(\dfrac{3}{4}\cdot\dfrac{7}{2}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{21}{8}+\dfrac{14}{3}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{175}{24}-\dfrac{1}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{851}{120}+\dfrac{17}{51}=\dfrac{297}{40}\)

`l )`

\(\dfrac{1}{3}\cdot3\dfrac{1}{2}-4\dfrac{2}{5}-\dfrac{26}{78}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\cdot\dfrac{7}{2}-\dfrac{22}{5}-\dfrac{1}{3}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\left(\dfrac{7}{2}-1\right)-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{1}{3}\cdot\dfrac{5}{2}-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(\dfrac{5}{6}-\dfrac{22}{5}+\dfrac{17}{51}\)

`=`\(-\dfrac{107}{30}+\dfrac{17}{51}=-\dfrac{97}{30}\)

P/s: Bạn tách bài ra hỏi nhé! Và ghi đề rõ ràng chứ đừng ghi ntnay, nhiều bạn nhìn vào rất khó nhìn!

`# \text {KaizulvG}`

GH
26 tháng 6 2023

24 tháng 8 2019

Bài làm

\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right]\)

\(=\left[\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right]:\left[\frac{25}{4002}.\frac{2001}{25}+\frac{9}{2}\right]\)

\(=\left[\frac{1}{34}+\frac{33}{34}\right]:\left[\frac{1}{2}+\frac{9}{2}\right]\)

\(=\frac{1}{5}\)

Bài 3 là hỗn số hả em?

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:\(x=\dfrac{15}{19}\)