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17 tháng 8 2023

=1,4 x\(\dfrac{15}{49}-\) \(\left(\dfrac{4}{5}+\dfrac{2}{3}\right)\) : 2\(\dfrac{1}{5}\)

=  \(\dfrac{3}{7}\) - \(\dfrac{22}{15}\) : \(\dfrac{11}{5}\)

\(\dfrac{3}{7}\) -  \(\dfrac{2}{3}\)

\(-\dfrac{5}{21}\) 

 

17 tháng 8 2023

( 2\(\dfrac{1}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(x\)) = \(\dfrac{3}{4}\)

  \(\dfrac{11}{5}\) + \(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\)

            \(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\) - \(\dfrac{11}{5}\)

           \(\dfrac{3}{5}\)\(x\) = - \(\dfrac{29}{20}\)

             \(x\) = -\(\dfrac{29}{12}\)

26 tháng 7 2015

-22/3

53/20

-240/7

6 tháng 3 2023

\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)

6 tháng 3 2023

fan liver tốt ghê

6 tháng 8 2018

\(1\frac{1}{2}+x=\frac{3}{2}-7\)

<=> \(\frac{3}{2}+x=\frac{-11}{2}\)

<=> \(x=-7\)

\(\frac{1}{4}+\frac{1}{3}:3x=-5\)

<=> \(\frac{1}{3}:3x=\frac{-21}{4}\)

<=> \(3x=\frac{-4}{63}\)

<=> \(x=\frac{4}{189}\)

\(\frac{4}{5}.x=\frac{8}{35}\)

<=> \(x=\frac{2}{7}\)

\(\frac{2}{3x}-\frac{1}{4}=\frac{7}{1}\)

<=> \(\frac{2}{3x}=\frac{29}{4}\)

=> \(8=87x\)

<=> \(x=\frac{8}{87}\)

\(\frac{3}{5x}+\frac{1}{2}=\frac{1}{7}\)

<=> \(\frac{3}{5x}=\frac{-5}{14}\)

<=> \(-25x=42\)

<=> \(x=\frac{-42}{25}\)

\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16.\frac{2}{3}\right)=0\)

<=> \(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{-32}{3}=0\)

<=> \(\frac{43}{8}+x-\frac{173}{24}=\frac{-32}{3}\)

<=> \(\frac{43}{8}+x=\frac{-83}{24}\)

<=> \(x=\frac{-53}{6}\)

học tốt

a: \(\Leftrightarrow\left|x\cdot\dfrac{7}{3}-\dfrac{3}{4}\right|=1+\dfrac{1}{3}+\dfrac{2}{3}=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{7}{3}-\dfrac{3}{4}=2\\x\cdot\dfrac{7}{3}-\dfrac{3}{4}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33}{28}\\x=-\dfrac{15}{28}\end{matrix}\right.\)

b: \(\Leftrightarrow\left|x\cdot\dfrac{2}{3}-\dfrac{1}{3}\right|=\dfrac{6}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{6}{5}\\x\cdot\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13}{10}\\x=\dfrac{23}{10}\end{matrix}\right.\)