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1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(\(x\) + \(\dfrac{1}{2}\))2 = \(\dfrac{1}{16}\)
\(\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- \(\dfrac{3}{4};-\dfrac{1}{4}\)}
\(x\) : (- \(\dfrac{1}{3}\))3 = - \(\dfrac{1}{3}\)
\(x\) = (-\(\dfrac{1}{3}\)).(-\(\dfrac{1}{3}\))3
\(x\) = \(\dfrac{1}{81}\)
Vậy \(x=\dfrac{1}{81}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(\Rightarrow3x-\frac{3}{2}-5x-3=-x+\frac{1}{5}\)
\(\Rightarrow3x-5x+x=\frac{1}{5}+\frac{3}{2}+3\)
\(\Rightarrow-x=\frac{17}{10}+3\)
\(\Rightarrow-x=\frac{47}{10}\)
\(\Rightarrow x=-\frac{47}{10}\)
\(\frac{1}{5}-\frac{1}{5-x}=\frac{1}{5}\)
\(\Rightarrow\frac{1}{5-x}=0\)
\(\Rightarrow5-x\in\varnothing\)
\(\Rightarrow x\in\varnothing\)
ban giup minh may cau nua duoc ko
thank bn