a) \(\dfrac{3}{4}+\dfrac{1}{4}\) : x \(=\) \(\dfrac{2}{5}\)
\(\dfrac{1}{4}\) : x \(=\) \(\dfrac{2}{5}\) \(-\) \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\) : x \(=\) \(\dfrac{8}{20}\) \(-\) \(\dfrac{15}{20}\)
\(\dfrac{1}{4}\) : x \(=\) \(\dfrac{-7}{20}\)
x \(=\) \(\dfrac{1}{4}\) : \(\dfrac{-7}{20}\)
x \(=\) \(\dfrac{1}{4}.\dfrac{-20}{7}\)
x \(=\) \(\dfrac{-5}{7}\)
Vậy x \(=\) \(\dfrac{-5}{7}\)
b) \(\dfrac{1}{4}+\dfrac{1}{3}\) : 2x \(=\) \(-5\)
\(\dfrac{1}{3}:2\)x \(=\) \(-5-\dfrac{1}{4}\)
\(\dfrac{1}{3}:2x\) \(=\) \(\dfrac{-20}{4}-\dfrac{1}{4}\)
\(\dfrac{1}{3}:2x\) \(=\) \(\dfrac{-21}{4}\)
2x \(=\) \(\dfrac{1}{3}:\dfrac{-21}{4}\)
2x \(=\) \(\dfrac{1}{3}.\dfrac{-4}{21}\)
2x \(=\) \(\dfrac{-4}{63}\)
x \(=\) \(\dfrac{-4}{63}:2\)
x \(=\) \(\dfrac{-4}{63}.\dfrac{1}{2}\)
x \(=\) \(\dfrac{-2}{63}\)
Vậy x \(=\) \(\dfrac{-2}{63}\)
c) \(-2\) \(+\) \(\left|x+\dfrac{1}{5}\right|\) \(=\dfrac{9}{10}\)
\(\left|x+\dfrac{1}{5}\right|\) \(=\) \(\dfrac{9}{10}\) \(-\left(-2\right)\)
\(\left|x+\dfrac{1}{5}\right|\) \(=\dfrac{9}{10}+2\)
\(\left|x+\dfrac{1}{5}\right|\) \(=\dfrac{9}{10}+\dfrac{20}{10}\)
\(\left|x+\dfrac{1}{5}\right|\) \(=\dfrac{29}{10}\)
\(\rightarrow\) \(\left\{{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{29}{10}\\x+\dfrac{1}{5}=\dfrac{-29}{10}\end{matrix}\right.\)
\(\rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{29}{10}-\dfrac{1}{5}\\x=\dfrac{-29}{10}-\dfrac{1}{5}\end{matrix}\right.\)
\(\rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{29}{10}-\dfrac{2}{10}\\x=\dfrac{-29}{10}-\dfrac{2}{10}\end{matrix}\right.\)
\(\rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{27}{10}=2,7\\x=\dfrac{-31}{10}=-3,1\end{matrix}\right.\)
Vậy x \(=\) 2,7 hoặc \(-3,1\)
a) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Leftrightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
\(a,\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}.\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}.\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}.\)
\(x=\dfrac{1}{4}:\left(-\dfrac{7}{20}\right).\)
\(x=\dfrac{1}{4}.\left(-\dfrac{20}{7}.\right)\)
\(x=-\dfrac{5}{7}.\)
Vậy.....
\(b,\dfrac{1}{4}+\dfrac{1}{3}:2x=-5.\)
\(\dfrac{1}{3}:2x=-5-\dfrac{1}{4}.\)
\(\dfrac{1}{3}:2x=-\dfrac{21}{4}.\)
\(2x=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right).\)
\(2x=\dfrac{1}{3}.\left(-\dfrac{4}{21}\right).\)
\(2x=-\dfrac{4}{63}.\)
\(x=-\dfrac{4}{63}:2.\)
\(x=-\dfrac{2}{63}.\)
Vậy.....
\(c,-2+\left|x+\dfrac{1}{5}\right|=\dfrac{9}{10}.\)
\(\left|x+\dfrac{1}{5}\right|=\dfrac{9}{10}-\left(-2\right).\)
\(\left|x+\dfrac{1}{5}\right|=\dfrac{29}{10}.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{29}{10}.\\x+\dfrac{1}{5}=-\dfrac{20}{10}.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{10}-\dfrac{1}{5}.\\x=-\dfrac{29}{10}-\dfrac{1}{5}.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{10}.\\x=-\dfrac{31}{10}.\end{matrix}\right.\)
Vậy.....
b) \(\dfrac{1}{4}+\dfrac{1}{3}:2x=-5\)
\(\Leftrightarrow\dfrac{1}{3}:2x=\left(-5\right)-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{3}:2x=\dfrac{-21}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{3}:\dfrac{-21}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{3}.\dfrac{4}{-21}\)
\(\Leftrightarrow2x=\dfrac{4}{-63}\)
\(\Leftrightarrow x=\dfrac{4}{-63}.\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{2}{63}\)