Bài 3:
a) Ta có: \(x-\dfrac{4}{15}=\dfrac{-3}{10}\)
\(\Leftrightarrow x=\dfrac{-3}{10}+\dfrac{4}{15}=\dfrac{-9}{30}+\dfrac{8}{30}\)
hay \(x=-\dfrac{1}{30}\)
Vậy: \(x=-\dfrac{1}{30}\)
c) Ta có: \(\left|x+1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-6\right\}\)
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