Bài 1:
a)
\(x-\dfrac{1}{15}=-\dfrac{1}{10}\\ \Rightarrow x=-\dfrac{1}{10}+\dfrac{1}{15}\\ \Rightarrow x=-\dfrac{1}{30}\)
b)
\(-x-\dfrac{1}{3}=-\dfrac{2}{5}\\ \Rightarrow-x=-\dfrac{2}{5}+\dfrac{1}{3}\\ \Rightarrow-x=-\dfrac{1}{15}\\ \Rightarrow x=\dfrac{1}{15}\)
c)
\(\left|2x-5\right|=4\\ \Rightarrow\left[{}\begin{matrix}2x-5=-4\\2x-5=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
Bài 2:
a) \(-\dfrac{6}{25}+\left|-\dfrac{4}{5}\right|-\left|\dfrac{2}{25}\right|\)
\(=\dfrac{-6}{25}+\dfrac{4}{5}-\dfrac{2}{25}\\ =\dfrac{-6}{25}+\dfrac{20}{25}-\dfrac{2}{25}\\ =\dfrac{12}{25}\)
b) \(\left(-\dfrac{5}{9}\right).\dfrac{3}{11}+\left(-\dfrac{13}{18}\right).\dfrac{3}{11}\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{5}{9}\right)+\left(-\dfrac{13}{18}\right)\right]\)
\(=\dfrac{3}{11}\left(-\dfrac{5}{9}-\dfrac{13}{18}\right)\)
\(=\dfrac{3}{11}\left(-\dfrac{10}{18}-\dfrac{13}{18}\right)\)
\(=\dfrac{3}{11}.\left(-\dfrac{23}{18}\right)\)
\(=-\dfrac{23}{66}\)
Bài 1:
a) Ta có: \(x-\dfrac{1}{15}=-\dfrac{1}{10}\)
nên \(x=\dfrac{-1}{10}+\dfrac{1}{15}\)
hay \(x=-\dfrac{1}{30}\)
b) Ta có: \(-x-\dfrac{1}{3}=-\dfrac{2}{5}\)
nên \(-x=\dfrac{-2}{5}+\dfrac{1}{3}=-\dfrac{1}{15}\)
hay \(x=\dfrac{1}{15}\)
c) Ta có: \(\left|2x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=4\\2x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
a) \(\dfrac{-6}{25}+\left|-\dfrac{4}{5}\right|-\left|\dfrac{2}{25}\right|\)
\(=\dfrac{-6}{25}+\dfrac{4}{5}-\dfrac{2}{25}\)
\(=\dfrac{-8}{25}+\dfrac{4}{5}=\dfrac{12}{25}\)
b) \(\left(-\dfrac{5}{9}\right)\cdot\dfrac{3}{11}+\left(-\dfrac{13}{18}\right)\cdot\dfrac{3}{11}\)
\(=\dfrac{3}{11}\cdot\left(-\dfrac{5}{9}-\dfrac{13}{18}\right)\)
\(=\dfrac{3}{11}\cdot\dfrac{-23}{18}=-\dfrac{23}{66}\)