a/ \(\left|x+\dfrac{1}{7}\right|=\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{7}=\dfrac{2}{3}\\x+\dfrac{1}{7}=\dfrac{-2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{21}\\x=\dfrac{-17}{21}\end{matrix}\right.\)
Vậy ................
b/ \(\left|x\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow\left|x\right|=\dfrac{7}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=\dfrac{-7}{4}\end{matrix}\right.\)
Vậy ...
c/ \(\dfrac{4}{5}-\left|x-\dfrac{1}{6}\right|=\dfrac{2}{3}\)
\(\Leftrightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{2}{15}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{2}{15}\\x-\dfrac{1}{6}=\dfrac{-2}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=\dfrac{1}{30}\end{matrix}\right.\)
Vậy ..
\(\left|x+\dfrac{1}{7}\right|=\dfrac{2}{3}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{7}=\dfrac{2}{3}\\x+\dfrac{1}{7}=\dfrac{-2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{21}\\x=\dfrac{-17}{21}\end{matrix}\right.\)
\(\left|x\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{1}{4}\\x-\dfrac{3}{2}=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=\dfrac{5}{4}\end{matrix}\right.\)
\(\dfrac{4}{5}-\left|x-\dfrac{1}{6}\right|=\dfrac{2}{3}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{4}{5}-\dfrac{2}{3}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{2}{15}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{2}{15}\\x-\dfrac{1}{6}=\dfrac{-2}{15}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=\dfrac{1}{30}\end{matrix}\right.\)
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