\(\frac{3}{x}\)=\(\frac{51}{85}\)

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3:x=51/85

3:x=3:5

x=5

8 tháng 2 2017

5

tk nha

11 tháng 2 2017

câu 1 a=85

câu 2\(\frac{7}{18}\)

câu 3 \(\frac{1}{18}\)

11 tháng 2 2017

Câu 1:a=85

Câu 2:x=7/18

Câu 3:x=1/18

9 tháng 8 2020

\(\frac{2}{3}.\frac{4}{7}=\frac{8}{21}\)

\(\frac{3}{11}.2=\frac{6}{11}\)

\(4.\frac{2}{7}=\frac{8}{7}\)

\(\frac{8}{21}:\frac{2}{3}=\frac{8}{21}.\frac{3}{2}=\frac{21}{2.21}=\frac{1}{2}\)

\(\frac{3}{7}.\frac{7}{3}=\frac{21}{21}=1\)

\(\frac{3}{7}:\frac{3}{7}=\frac{3}{7}.\frac{7}{3}=\frac{21}{21}=1\)

9 tháng 8 2020

lỡ tay bấm gửi trả lời luôn

\(\frac{2}{3}.\frac{1}{6}.\frac{9}{11}=\frac{2.9}{18.11}=\frac{2.9}{2.9.11}=\frac{1}{11}\)

\(\frac{2.3.4}{2.3.4.5}=\frac{6.4}{6.4.5}=\frac{24}{24.5}=\frac{1}{5}\)

1 tháng 6 2017

\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)

\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)

\(=\frac{1}{24}\)

1 tháng 6 2017

\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)

\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)

\(=\frac{1}{10}\)

4 tháng 3 2019

1) \(\frac{4}{5}.\frac{8}{3}.\frac{x}{7}=\frac{96}{105}\)

\(\frac{32.x}{105}=\frac{96}{105}\)

\(32x=96\)

\(x=3\)

9 tháng 9 2018

\(x+\frac{2}{3}=\frac{9}{11}\)

\(x=\frac{9}{11}-\frac{2}{3}\)

\(x=\frac{5}{33}\)

\(x-\frac{3}{10}=\frac{4}{15}\)

\(x=\frac{3}{10}-\frac{4}{15}\)

\(x=\frac{1}{10}\)

\(x\times\frac{1}{7}=\frac{5}{6}\)

\(x=\frac{5}{6}:\frac{1}{7}\)

\(x=\frac{35}{6}\)

\(x:\frac{3}{5}=\frac{1}{6}\)

\(x=\frac{3}{5}.\frac{1}{6}\)

\(x=\frac{1}{10}\)

+, \(x+\frac{2}{3}=\frac{9}{11}\)

\(x=\frac{9}{11}-\frac{2}{3}=\frac{5}{33}\)

+, \(x-\frac{3}{10}=\frac{4}{15}\)

\(x=\frac{4}{15}+\frac{3}{10}=\frac{17}{30}\)

+, \(x\cdot\frac{1}{7}=\frac{5}{6}\)

\(x=\frac{5}{6}:\frac{1}{7}=\frac{5}{6}.\frac{7}{1}=\frac{35}{6}\)

+,\(x:\frac{3}{5}=\frac{1}{6}\)

\(x=\frac{1}{6}\cdot\frac{3}{5}=\frac{3}{30}=\frac{1}{10}\)

16 tháng 6 2020

\(a.\frac{4}{3}-\frac{3}{2}:X=\frac{1}{6}\)

           \(\frac{3}{2}:X=\frac{4}{3}-\frac{1}{6}\)

           \(\frac{3}{2}:X=\frac{8}{6}-\frac{1}{6}\)

           \(\frac{3}{2}:X=\frac{7}{6}\)

               \(X=\frac{3}{2}:\frac{7}{6}\)

              \(X=\frac{3}{2}\times\frac{6}{7}\)

              \(X=\frac{9}{7}\)

16 tháng 6 2020

\(b.\left(X+\frac{2}{3}\right):\frac{1}{3}=\frac{41}{3}\)

    \(X-\frac{2}{3}=\frac{41}{3}.\frac{1}{3}\)

   \(X-\frac{2}{3}=\frac{41}{9}\)

  \(X=\frac{41}{9}+\frac{2}{3}\)

  \(X=\frac{41}{9}+\frac{6}{9}\)

 \(X=\frac{47}{9}\)

7 tháng 7 2018

a) \(\frac{3}{5}\times y+\frac{1}{2}:\frac{5}{3}-\frac{5}{4}=\frac{1}{2}\times\frac{1}{3}\)

\(\Rightarrow\frac{3}{5}\times y+\frac{3}{10}-\frac{5}{4}=\frac{1}{6}\)

\(\Rightarrow\frac{3}{5}\times y+\left(-\frac{19}{20}\right)=\frac{1}{6}\)

\(\Rightarrow\frac{3}{5}\times y=\frac{67}{60}\)

\(\Rightarrow y=\frac{67}{36}\)

b) \(\frac{4}{5}:y+\frac{1}{4}\times\frac{1}{6}-\frac{1}{2}=\frac{1}{3}\times\frac{5}{2}\)

\(\Rightarrow\frac{4}{5}:y+\frac{1}{24}-\frac{1}{2}=\frac{5}{6}\)

\(\Rightarrow\frac{4}{5}:y+\left(-\frac{11}{24}\right)=\frac{5}{6}\)

\(\Rightarrow\frac{4}{5}:y=\frac{5}{6}+\frac{11}{24}=\frac{31}{24}\)

\(\Rightarrow y=\frac{4}{5}:\frac{31}{24}=\frac{96}{155}\)

c) \(\frac{3}{5}\times y-\frac{4}{5}:3+\frac{1}{12}=\frac{3}{2}+\frac{1}{5}\)

\(\Rightarrow\frac{3}{5}\times y-\frac{4}{15}+\frac{1}{12}=\frac{17}{10}\)

\(\Rightarrow\frac{3}{5}\times y-\frac{4}{15}=\frac{97}{60}\)

\(\Rightarrow\frac{3}{5}\times y=\frac{113}{60}\)

\(\Rightarrow y=\frac{113}{36}\)

13 tháng 9 2021

\(\frac{90}{100}=\frac{90:10}{100:10}=\frac{9}{10}\)

\(\frac{64}{128}=\frac{64:64}{128:64}=\frac{1}{2}\)

\(\frac{17}{51}=\frac{17:17}{51:17}=\frac{1}{3}\)

\(\frac{152}{608}=\frac{152:152}{608:152}=\frac{1}{4}\)

\(\frac{63}{315}=\frac{63:63}{315:63}=\frac{1}{5}\)