Question

Answer 1

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: \(x\in\left\{4;-6\right\}\)