\(\left|2x-\dfrac{1}{2}\right|+\dfrac{3}{7}=5\dfrac{3}{7}\)
<=> \(\left|2x-\dfrac{1}{2}\right|=5\)
<=> \(\left\{{}\begin{matrix}2x-\dfrac{1}{2}=5\\2x-\dfrac{1}{2}=-5\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}2x=\dfrac{11}{2}\\2x=-\dfrac{9}{2}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=\dfrac{11}{4}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Do x < 0 => x = \(-\dfrac{9}{4}\)
Giải:
\(\left|2x-\dfrac{1}{2}\right|+\dfrac{3}{7}=5\dfrac{3}{7}\)
\(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|=5\dfrac{3}{7}-\dfrac{3}{7}\)
\(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=4\\2x-\dfrac{1}{2}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{9}{2}\\2x=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
Mà x < 0
Nên \(x=-\dfrac{7}{4}\)
Vậy \(x=-\dfrac{7}{4}\).
Chúc bạn học tốt!
\(\left|2x-\dfrac{1}{2}\right|+\dfrac{3}{7}=5\dfrac{3}{7}\)
\(\Rightarrow\left|2x-\dfrac{1}{2}\right|=5\dfrac{3}{7}-\dfrac{3}{7}\)
\(\Rightarrow\left|2x-\dfrac{1}{2}\right|=\dfrac{38}{7}-\dfrac{3}{7}\)
\(\Rightarrow\left|2x-\dfrac{1}{2}\right|=5\)
\(\Rightarrow2x-\dfrac{1}{2}=\left[{}\begin{matrix}5\\-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=5+\dfrac{1}{2}\\2x=\left(-5\right)+\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{11}{2}\\2x=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\dfrac{11}{2}}{2}\\x=\dfrac{\dfrac{-9}{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
